A1001(20 两数相加)

#include<iostream>
#include<string>
using namespace std;
int main(){int a,b;cin>>a>>b;string s=to_string(a+b);int len=s.length();for(int i=0;i<len;i++){cout<<s[i];if(s[i]=='-') continue;if((i+1)%3==len%3&&i!=len-1) cout<<",";}return 0;
}

A1002(多项式求和)

#include<cstdio>
#include<string.h>
const int maxn=1001;
int count1,count2;
double coe[maxn]={},a;
int number=0;
int main(){scanf("%d",&count1);while(count1--){int i;scanf("%d%lf",&i,&a);coe[i]=coe[i]+a;}scanf("%d",&count2);while(count2--){int k;scanf("%d%lf",&k,&a);coe[k]=coe[k]+a;}for(int k=maxn-1;k>=0;k--){if(coe[k]!=0){number++;}}printf("%d",number);for(int k=maxn-1;k>=0;k--){if(coe[k]!=0) printf(" %d %.1f",k,coe[k]);}return 0;
}

A1003(25 Dijkstra、点权、边权)

#include<bits/stdc++.h>
using namespace std;
const int maxn=510;
const int INF=0x7fffffff;
int n,m,c1,c2,weight[maxn],g[maxn][maxn],d[maxn],w[maxn],num[maxn]={0};
bool vis[maxn]={false};
void dijkst(int s){fill(d,d+maxn,INF);fill(vis,vis+maxn,false);fill(w,w+maxn,0);d[s]=0;num[s]=1;w[s]=weight[s];for(int i=0;i<n;i++){int u=-1,MIN=INF;for(int j=0;j<n;j++){if(vis[j]==false&&d[j]<MIN){u=j;MIN=d[j];}}if(u==-1) return;vis[u]=true;for(int j=0;j<n;j++){if(g[u][j]!=INF&&vis[j]==false){if(d[u]+g[u][j]<d[j]){d[j]=d[u]+g[u][j];num[j]=num[u];w[j]=w[u]+weight[j];}else if(d[u]+g[u][j]==d[j]){num[j]+=num[u];if(w[u]+weight[j]>w[j]) w[j]=w[u]+weight[j];}}}}
}
int main(){cin>>n>>m>>c1>>c2;for(int i=0;i<n;i++) cin>>weight[i];fill(g[0],g[0]+maxn*maxn,INF);for(int i=0;i<m;i++){int v1,v2,t;cin>>v1>>v2>>t;g[v1][v2]=g[v2][v1]=t;}dijkst(c1);printf("%d %d",num[c2],w[c2]);return 0;
}

A1004

1.DFS法
#include<cstdio>
#include<vector>
using namespace std;
int n,m,level[100]={0},maxdepth=-1;
vector<int> tree[100];
void dfs(int index,int depth){if(tree[index].size()==0){level[depth]++;if(depth>maxdepth)    maxdepth=depth;return;}for(int i=0;i<tree[index].size();i++) dfs(tree[index][i],depth+1);
}
int main(){int id,k,temp;scanf("%d%d",&n,&m);for(int i=0;i<m;i++){scanf("%d%d",&id,&k);while(k--){scanf("%d",&temp);tree[id].push_back(temp);}}dfs(1,0);for(int i=0;i<=maxdepth;i++){if(i>0) printf(" ");printf("%d",level[i]);}return 0;
}2.BFS法
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 110;
int hashtable[maxn] = { 0 };
vector<int> child[maxn];
int n, m, level[maxn] = { 0 },maxlevel=0;
void BFS() {level[1] = 1;queue<int> q;q.push(1);while (!q.empty()) {int now = q.front();q.pop();if (child[now].size() == 0) {hashtable[level[now]]++;maxlevel = max(maxlevel, level[now]);}for (int i = 0; i < child[now].size(); i++) {level[child[now][i]] = level[now] + 1;q.push(child[now][i]);}}
}
int main() {scanf("%d%d", &n, &m);int father,num,temp;while (m--) {scanf("%d%d", &father, &num);while (num--) {scanf("%d", &temp);child[father].push_back(temp);}}BFS();for (int i = 1; i <= maxlevel; i++) {printf("%d", hashtable[i]);if (i < maxlevel) printf(" ");}return 0;
}

A1005(20 字符串)

#include<iostream>
#include<string>
using namespace std;
int main(){string n,spell[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};cin>>n;int sum=0;for(int i=0;i<n.size();i++) sum+=(n[i]-'0');string ans=to_string(sum);cout<<spell[ans[0]-'0'];for(int i=1;i<ans.size();i++) cout<<" "<<spell[ans[i]-'0'];return 0;
}

A1006

#include<cstdio> struct person {char id[16];int hh, mm, ss;
}temp,first,last;
bool great(person node1, person node2) {if (node1.hh != node2.hh) return node1.hh > node2.hh;if (node2.mm != node2.mm) return node1.mm > node2.mm;return node1.ss > node2.ss;
}
int main() {int n;scanf("%d", &n);first.hh = 24, first.mm = 60, first.ss = 60;last.hh = 0, last.mm = 0, last.ss = 0;for (int i = 0; i < n; i++) {scanf("%s %d:%d:%d", temp.id, &temp.hh, &temp.mm, &temp.ss);if (great(temp, first) == false)  first = temp;scanf("%d:%d:%d", &temp.hh, &temp.mm, &temp.ss);if (great(temp, last))   last = temp;}printf("%s %s", first.id, last.id);return 0;
}

A1007

#include<cstdio>
const int maxn=10010;
int a[maxn],dp[maxn],s[maxn]={0};
int main(){int n;scanf("%d",&n);bool flag=false;for(int i=0;i<n;i++){scanf("%d",&a[i]);if(a[i]>=0) flag=true;}if(flag==false){printf("0 %d %d",a[0],a[n-1]);return 0;}dp[0]=a[0];for(int i=1;i<n;i++){if(a[i]>dp[i-1]+a[i]){dp[i]=a[i];s[i]=i;} else {dp[i]=dp[i-1]+a[i];s[i]=s[i-1];}}int k=0;for(int i=1;i<n;i++){if(dp[i]>dp[k]){k=i;}}printf("%d %d %d",dp[k],a[s[k]],a[k]);return 0;
}

A1008(20 水题)

#include<iostream>
#include<vector>
using namespace std;
int main(){int n,ans=0;cin>>n;vector<int> v(n+1,0);for(int i=1;i<=n;i++){cin>>v[i];if(v[i]>v[i-1]) ans+=(v[i]-v[i-1])*6;else ans+=(v[i-1]-v[i])*4;ans+=5;}cout<<ans;return 0;
}

A1009 多项式相乘

#include<cstdio>
#include<string.h>
const int maxn=2001;
int count1,count2;
struct poly{int exp;double coef;
}coe[1001];
double result[maxn];
int number=0;
int main(){scanf("%d",&count1);
for(int i =0;i<count1;i++){scanf("%d%lf",&coe[i].exp,&coe[i].coef);}scanf("%d",&count2);
for(int i=0;i<count2;i++){int k;double a;scanf("%d%lf",&k,&a);for(int i=0;i<count1;i++){result[k+coe[i].exp]+=(a*coe[i].coef);}}for(int k=maxn-1;k>=0;k--){if(result[k]!=0){number++;}}printf("%d",number);for(int k=maxn-1;k>=0;k--){if(result[k]!=0) printf(" %d %.1f",k,result[k]);}return 0;
}

A1010(25 二分)

#include<iostream>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
long long convert(string n,long long radix){long long sum=0,index=0,temp;for(auto it=n.rbegin();it!=n.rend();it++){temp=isdigit(*it)? (*it)-'0' : (*it)-'a'+10;sum+=temp*pow(radix,index++);}return sum;
}
long long findradix(string n,long long num){char it=*max_element(n.begin(),n.end());long long low=(isdigit(it)? (it)-'0' : (it)-'a'+10)+1;long long high=max(low,num);while(low<=high){long long mid=(low+high)/2;long long temp=convert(n,mid);if(temp<0||temp>num) high=mid-1;//小于0的情况是进制太大导致溢出 else if(temp==num) return mid;else low=mid+1;}return -1;
}
int main(){string n1,n2;long long tag,radix,ans;cin>>n1>>n2>>tag>>radix;ans=tag==1?findradix(n2,convert(n1,radix)) : findradix(n1,convert(n2,radix));if(ans==-1) cout<<"Impossible";else cout<<ans;return 0;
}

A1011

#include<cstdio>
char s[4] = "WTL";
double a,res=1;
int main() {for (int i = 0; i < 3; i++) {int imax;double temp = 0;for (int i = 0; i < 3; i++) {scanf_s("%lf", &a);if (a > temp) {temp = a;imax = i;}}res *= temp;printf("%c ", s[imax]);}printf("%.2f", (res * 0.65 - 1) * 2);return 0;
}

A1012(25 条件排序)

#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
struct node{int id;int score[4];
}stu[2000];
int now,rk[1000000][4];
bool cmp(node a,node b){return a.score[now]>b.score[now];
}
int main(){int n,m;scanf("%d%d",&n,&m);map<int,bool> mp;for(int i=0;i<n;i++){scanf("%d%d%d%d",&stu[i].id,&stu[i].score[1],&stu[i].score[2],&stu[i].score[3]);stu[i].score[0]=(int)((stu[i].score[1]+stu[i].score[2]+stu[i].score[3])*1.0/3+0.5);mp[stu[i].id]=true;}for(now=0;now<=3;now++){sort(stu,stu+n,cmp);rk[stu[0].id][now]=1;for(int i=1;i<n;i++){if(stu[i].score[now]==stu[i-1].score[now]) rk[stu[i].id][now]=rk[stu[i-1].id][now];else rk[stu[i].id][now]=i+1;}}char course[4]={'A','C','M','E'};for(int i=0;i<m;i++){int query,bestrank=9999,index=0;scanf("%d",&query);if(mp.count(query)==0){printf("N/A\n");continue;}for(int j=0;j<4;j++){if(rk[query][j]<bestrank){bestrank=rk[query][j];index=j;}}printf("%d %c\n",rk[query][index],course[index]);}return 0;
}

A1013(25 图的dfs遍历，计算连通块个数)

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=10010;
int n,m,k,G[maxn][maxn],cnt=0;
bool vis[maxn];
void dfs(int v){vis[v]=true;for(int i=1;i<=n;i++){if(G[v][i]==1&&vis[i]==false) dfs(i);}
}
int main(){int a,b;scanf("%d%d%d",&n,&m,&k);while(m--){scanf("%d%d",&a,&b);G[a][b]=G[b][a]=1;}while(k--){int cnt=0;fill(vis,vis+maxn,false);scanf("%d",&a);vis[a]=true;for(int i=1;i<=n;i++){if(vis[i]==false){dfs(i);cnt++;}}printf("%d\n",cnt-1);//需要添加的路径为连通块个数减1 }return 0;
}

A1014(30 模拟 难题)

#include<vector>
#include<cstdio>
#include<queue>
using namespace std;
struct node {int poptime, endtime;queue<int> q;
};
int main() {int n, m, k, Q, index = 1, query;scanf("%d%d%d%d", &n, &m, &k, &Q);vector<int> time(k + 1), ans(k + 1);for (int i = 1; i <= k; i++) scanf("%d", &time[i]);vector<node>window(n + 1);vector<bool>sorry(k + 1, false);for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (index <= k) {window[j].q.push(time[index]);if (window[j].endtime >= 540) sorry[index] = true;window[j].endtime += time[index];if (i == 1) window[j].poptime = window[j].endtime;ans[index] = window[j].endtime;index++;}}}while (index <= k) {int minpoptime = window[1].poptime, tempwindow = 1;for (int i = 2; i <= n; i++) {if (window[i].poptime < minpoptime) {minpoptime = window[i].poptime;tempwindow = i;}}window[tempwindow].q.pop();window[tempwindow].poptime += window[tempwindow].q.front();window[tempwindow].q.push(time[index]);if (window[tempwindow].endtime >= 540) sorry[index] = true;window[tempwindow].endtime += time[index];ans[index] = window[tempwindow].endtime;index++;}while (Q--) {scanf("%d", &query);if (sorry[query] == true) printf("Sorry\n");else printf("%02d:%02d\n", (ans[query] + 480) / 60, (ans[query] + 480) % 60);}return 0;
}

A1015(20 进制转化与素数判断)

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
int change(int n,int b){vector<int> v;int ans=0;while(n!=0){v.push_back(n%b);n/=b;}for(int i=v.size()-1;i>=0;i--) ans+=v[i]*pow(b,v.size()-1-i);return ans;
}
bool isprime(int n){if(n<=1) return false;int sqr=sqrt(n*1.0);for(int i=2;i<=sqr;i++){if(n%i==0) return false;}return true;
}
int main(){int n,b;while(1){scanf("%d",&n);if(n<0) return 0;scanf("%d",&b);if(isprime(n)&&isprime(change(n,b))) printf("Yes\n");else printf("No\n");}
}

A1016(25 排序 处理多条通话时间记录)

#include<iostream>
#include<algorithm>
#include<map>
#include<vector>
#include<string>
using namespace std;
struct node{string name;int status,time,month,day,hour,minute;
};
bool cmp(node a,node b){return a.name!=b.name? a.name<b.name : a.time<b.time;
}
double billfromzero(node a,int *rate){double total=a.minute*rate[a.hour]+a.day*60*rate[24];for(int i=0;i<a.hour;i++) total+=60*rate[i];return total/100.0;
}int main(){int n,rate[25]={0};for(int i=0;i<24;i++){scanf("%d",&rate[i]);rate[24]+=rate[i];}cin>>n;vector<node> v(n);for(int i=0;i<n;i++){cin>>v[i].name;scanf("%d:%d:%d:%d",&v[i].month,&v[i].day,&v[i].hour,&v[i].minute);v[i].time=v[i].day*24*60+v[i].hour*60+v[i].minute;string temp;cin>>temp;v[i].status=(temp=="on-line")? 1 : 0 ;} sort(v.begin(),v.end(),cmp);map<string,vector<node> >mp;for(int i=1;i<n;i++){if(v[i].name==v[i-1].name&&v[i-1].status==1&&v[i].status==0){mp[v[i-1].name].push_back(v[i-1]);mp[v[i].name].push_back(v[i]);}}for(auto it:mp){cout<< it.first;vector<node> temp=it.second;printf(" %02d\n",temp[0].month);double total=0.0;for(int i=1;i<temp.size();i+=2){double t=billfromzero(temp[i],rate)-billfromzero(temp[i-1],rate);printf("%02d:%02d:%02d",temp[i-1].day,temp[i-1].hour,temp[i-1].minute);printf(" %02d:%02d:%02d %d",temp[i].day,temp[i].hour,temp[i].minute,temp[i].time-temp[i-1].time);printf(" $%.2f\n",t);total+=t;}printf("Total amount: $%.2f\n",total);} return 0;}

A1017 （25 模拟 ，时间处理、排序、）

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{int cometime,servetime;
};
bool cmp(node a,node b){return a.cometime<b.cometime;
}
int main(){int n,k;scanf("%d%d",&n,&k);double ans=0.0;node temp;vector<node> v;for(int i=0;i<n;i++){int hh,mm,ss,serve;scanf("%d:%d:%d %d",&hh,&mm,&ss,&serve);temp.cometime=hh*3600+mm*60+ss;temp.servetime=serve*60;if(temp.cometime>61200) continue;v.push_back(temp);}sort(v.begin(),v.end(),cmp);vector<int> window(k,28800);for(int i=0;i<v.size();i++){int minend=window[0],index=0;for(int j=1;j<k;j++){if(window[j]<minend){minend=window[j];index=j;}}if(v[i].cometime>window[index]) window[index]=v[i].cometime+v[i].servetime;else if(v[i].cometime<=window[index]){ans+=(window[index]-v[i].cometime);window[index]+=v[i].servetime; }}if(v.size()==0) printf("0.0");else printf("%.1f",(ans/60.0)/v.size());return 0;
}

A1018(30 dijkst+dfs)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=520;
const int INF=0x7fffffff;
int cmax,n,sp,m,minneed=INF,minremain=INF;
int weight[maxn],G[maxn][maxn],d[maxn];
bool vis[maxn]={false};
vector<int>pre[maxn];
vector<int>temp,path;
void dijkst(int s){fill(d,d+maxn,INF);d[s]=0;for(int i=0;i<=n;i++){int u=-1,MIN=INF;for(int j=0;j<=n;j++){if(vis[j]==false&&d[j]<MIN){MIN=d[j];u=j;}}if(u==-1) return;vis[u]=true;for(int j=0;j<=n;j++){if(G[u][j]!=INF&&vis[j]==false){if(G[u][j]+d[u]<d[j]){d[j]=G[u][j]+d[u];pre[j].clear();pre[j].push_back(u);}else if(G[u][j]+d[u]==d[j]) pre[j].push_back(u);}}}
}
void dfs(int s){if(s==0){temp.push_back(s);//注意 int need=0,remain=0;for(int i=temp.size()-1;i>=0;i--){if(weight[temp[i]]>0) remain+=weight[temp[i]];if(weight[temp[i]]<0){if(abs(weight[temp[i]])>remain){need+=abs(weight[temp[i]])-remain;remain=0;}else remain-=abs(weight[temp[i]]);}}if(need<minneed){minneed=need;minremain=remain;//注意path=temp;}else if(need==minneed&&remain<minremain){minremain=remain;path=temp;}temp.pop_back();//注意return;}temp.push_back(s);for(int i=0;i<pre[s].size();i++) dfs(pre[s][i]);temp.pop_back();
}
int main(){scanf("%d%d%d%d",&cmax,&n,&sp,&m);for(int i=1;i<=n;i++) {scanf("%d",&weight[i]);weight[i]-=cmax/2;}fill(G[0],G[0]+maxn*maxn,INF);for(int i=0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);G[a][b]=G[b][a]=c;}dijkst(0);dfs(sp);printf("%d ",minneed);for(int i=path.size()-1;i>=0;i--){printf("%d",path[i]);if(i>0) printf("->");}printf(" %d",minremain);return 0;
}

A1019

#include<cstdio>int main(){int n,b,ans[50],index=0;scanf("%d%d",&n,&b);while(n!=0){ans[index++]=n%b;n/=b;}bool flag=true;for(int i=0;i<=index/2;i++){if(ans[i]!=ans[index-i-1]) flag=false;}if(flag==false) printf("No\n");else printf("Yes\n");for(int i=index-1;i>=0;i--){if(i!=index-1) printf(" ");printf("%d",ans[i]);}return 0; }

A1020(25 后序+中序建树，输出层序)

#include<cstdio>
#include<queue>
using namespace std;
const int maxn=31;
struct node{int data;node *left,*right;
};
int post[maxn],in[maxn],n,num=0;
node * create(int postl,int postr,int inl,int inr){if(postl>postr) return NULL;node *root=new node;root->data=post[postr];int k=inl;while(in[k]!=root->data) k++;int leftnum=k-inl;root->left=create(postl,postl+leftnum-1,inl,k-1);root->right=create(postl+leftnum,postr-1,k+1,inr);return root;
}
void layerorder(node *root){queue<node*> q;q.push(root);while(!q.empty()){node* Top=q.front();q.pop();printf("%d",Top->data);num++;if(num<n) printf(" ");if(Top->left!=NULL) q.push(Top->left);if(Top->right!=NULL) q.push(Top->right);}
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&post[i]);for(int i=0;i<n;i++) scanf("%d",&in[i]);node*root=create(0,n-1,0,n-1);layerorder(root);return 0;
}

A1021(25 找出图中使深度最深的结点(首先判断是否能看成树))

#include<cstdio>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
bool vis[10010];
int n,maxheight=-1;
vector<int> ans,G[10010];
void dfs(int index,int height){vis[index]=true;if(height>maxheight){maxheight=height;ans.clear();ans.push_back(index);}else if(height==maxheight) ans.push_back(index);for(int i=0;i<G[index].size();i++){if(vis[G[index][i]]==false) dfs(G[index][i],height+1); }
}
int main(){int v1,v2,cnt=0,st;set<int> res;scanf("%d",&n);for(int i=1;i<=n-1;i++){scanf("%d%d",&v1,&v2);G[v1].push_back(v2);G[v2].push_back(v1);} fill(vis,vis+10010,false);for(int i=1;i<=n;i++){if(vis[i]==false){dfs(i,0);if(i==1){if(ans.size()!=0) st=ans[0];for(int j=0;j<ans.size();j++) res.insert(ans[j]);} cnt++;}}if(cnt>1) printf("Error: %d components",cnt);else{ans.clear(); fill(vis,vis+10010,false);maxheight=-1;dfs(st,0);for(int i=0;i<ans.size();i++) res.insert(ans[i]);for(auto it=res.begin();it!=res.end();it++) printf("%d\n",*it);}return 0;
}

A1022

#include<cstdio>
#include<iostream>
#include<map>
#include<set>
#include<string>
using namespace std;
map<string, set<int> > mptitle, mpautor, mpkey, mppublish, mpyear;
void query(map<string, set<int> >& mp, string& str) {if (mp.count(str) == 0) cout << "Not Found\n";else {for (auto it = mp[str].begin(); it != mp[str].end(); it++) {printf("%07d\n", *it);}}
}
int main() {int n,m,id;cin >> n;string title, autor, keyword, publish, year;for (int i = 0; i < n; i++) {cin >> id;char c = getchar();getline(cin, title);mptitle[title].insert(id);getline(cin, autor);mpautor[autor].insert(id);while (cin >> keyword) {mpkey[keyword].insert(id);c = getchar();if (c == '\n') break;}getline(cin, publish);mppublish[publish].insert(id);getline(cin, year);mpyear[year].insert(id);}cin >> m;while(m--){int type;string temp;scanf("%d: ", &type);getline(cin, temp);cout << type << ":" << " " << temp<<endl;if (type == 1) query(mptitle, temp);if (type == 2) query(mpautor, temp);if (type == 3) query(mpkey, temp);if (type == 4) query(mppublish, temp);if (type == 5) query(mpyear, temp);}return 0;
}

A1023(20 大整数乘2)

方法一:
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
struct bign {int d[1000];int len;bign() {len = 0;memset(d, 0, sizeof(d));}
};
bign change(string str) {bign a;a.len = str.length();for (int i = 0; i < a.len; i++) {a.d[i] = str[a.len - 1 - i]-'0';}return a;
}
bign multi(bign a, int b) {bign c;int carry=0;for (int i = 0; i < a.len; i++) {int temp = a.d[i] * b + carry;c.d[c.len++] = temp % 10;carry = temp / 10;}while (carry != 0) {c.d[c.len++] = carry % 10;carry /= 10;}return c;
}
bool judge(bign a, bign b) {if (a.len != b.len) return false;else {int count[10] = { 0 };for (int i = 0; i < a.len; i++) {count[a.d[i]]++;count[b.d[i]]--;}for (int i = 0; i < 10; i++) {if (count[i] != 0) {return false;}}}return true;
}
void print(bign a) {for (int i = a.len - 1; i >= 0; i--) {printf("%d", a.d[i]);}
}
int main() {string str;cin >> str;bign a, ans;a=change(str);ans = multi(a, 2);if (judge(a, ans)) cout << "Yes" << endl;else cout << "No" << endl;print(ans);return 0;
}方法二:
#include<iostream>
#include<string>
#include<map>
using namespace std;
string Double(string s){string ans=s;int carry=0;for(int i=s.size()-1;i>=0;i--){ans[i]=(((s[i]-'0')*2+carry)%10)+'0';carry=((s[i]-'0')*2+carry)/10;}if(carry>0) ans="1"+ans;return ans;
}
int main(){string s,res;map<char,int>mp;cin>>s;for(int i=0;i<s.size();i++) mp[s[i]]++;res=Double(s);for(int i=0;i<res.size();i++) mp[res[i]]--;for(auto it=mp.begin();it!=mp.end();it++){if(it->second!=0){cout<<"No\n"<<res;return 0;}}cout<<"Yes\n"<<res;return 0;
}

A1024(25 大整数反转相加，判断回文数)

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string rev(string s) {reverse(s.begin(),s.end());return s;
}
string add(string s1,string s2) {string ans=s1;int carry=0;for(int i=ans.size()-1; i>=0; i--) {ans[i]=((s1[i]-'0')+(s2[i]-'0')+carry)%10+'0';carry=((s1[i]-'0')+(s2[i]-'0')+carry)/10;}if(carry!=0) ans="1"+ans;return ans;
}
bool judge(string s) {for(int i=0; i<s.size()/2; i++) {if(s[i]!=s[s.size()-i-1]) return false;}return true;
}
int main() {string s;int step;cin>>s>>step;if(s.size()==1||judge(s)) cout<<s<<endl<<"0";else {for(int i=1; i<=step; i++) {s=add(s,rev(s));if(judge(s)||i==step) {cout<<s<<endl<<i;break;}}}return 0;
}

A1025(25 条件排序)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct node{string id;int localid,score,frank,lrank;
};
bool cmp(node a,node b){return a.score!=b.score? a.score>b.score : a.id<b.id;
}
int main(){int n,k;cin>>n;vector<node> total;for(int j=1;j<=n;j++){scanf("%d",&k);vector<node> v(k);for(int i=0;i<k;i++) {cin>>v[i].id>>v[i].score;v[i].localid=j;}sort(v.begin(),v.end(),cmp);v[0].lrank=1;for(int i=1;i<k;i++) {if(v[i].score==v[i-1].score) v[i].lrank=v[i-1].lrank;else v[i].lrank=i+1;}for(int i=0;i<k;i++) total.push_back(v[i]);}sort(total.begin(),total.end(),cmp);total[0].frank=1;for(int i=1;i<total.size();i++) {if(total[i].score==total[i-1].score) total[i].frank=total[i-1].frank;else total[i].frank=i+1;}cout<<total.size()<<endl;for(int i=0;i<total.size();i++){printf("%s %d %d %d\n",total[i].id.c_str(),total[i].frank,total[i].localid,total[i].lrank);}return 0;
}

A1027

#include<cstdio>
int main(){char c[14]={"0123456789ABC"};printf("#");for(int i=0;i<3;i++){int n;scanf("%d",&n);printf("%c%c",c[n/13],c[n%13]);}return 0;
}

A1028(25 排序)

#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
using namespace std;
struct node{int id,grade;string name;
};
bool cmp1(node a,node b){return a.id<b.id;
}
bool cmp2(node a,node b){return a.name!=b.name? a.name<b.name :a.id<b.id;
}
bool cmp3(node a,node b){return a.grade!=b.grade? a.grade<b.grade : a.id<b.id;
}
int main(){int n,c;cin>>n>>c;vector<node> v(n);for(int i=0;i<n;i++) cin>>v[i].id>>v[i].name>>v[i].grade;if(c==1) sort(v.begin(),v.end(),cmp1);else if(c==2) sort(v.begin(),v.end(),cmp2);else if(c==3)sort(v.begin(),v.end(),cmp3);for(int i=0;i<n;i++) printf("%06d %s %d\n",v[i].id,v[i].name.c_str(),v[i].grade);return 0;
}

A1029(25 求中位数 twopoints)

1.直接暴力
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=400010;
int main(){int n1,n2;long long a[maxn];scanf("%d",&n1);int i;for( i=0;i<n1;i++){scanf("%lld",&a[i]);}scanf("%d",&n2);for( i=n1;i<n2+n1;i++){scanf("%lld",&a[i]);}sort(a,a+n1+n2);printf("%d",a[(n1+n2-1)/2]);return 0;
}2.two points#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=400010;
const int INF=(1<<31)-1;
int main(){int n1,n2;long long a[maxn],b[maxn];scanf("%d",&n1);for(int i=0;i<n1;i++){scanf("%lld",&a[i]);}scanf("%d",&n2);for(int i=0;i<n2;i++){scanf("%lld",&b[i]);}a[n1]=INF;b[n2]=INF;int midpos=(n1+n2-1)/2;int count=0,j=0,i=0;while(count<midpos){if(a[i]<b[j]) i++;else j++;count++;}if(a[i]<b[j]) printf("%lld",a[i]);else printf("%lld",b[j]);return 0;
}

A1030(30 dijkst,距离最短加花费最少)

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=510;
const int INF=0x7fffffff;
int G[maxn][maxn],n,m,st,ed,cost[maxn][maxn],d[maxn],c[maxn],pre[maxn];
bool vis[maxn];
void dijkst(int s){fill(d,d+maxn,INF);fill(c,c+maxn,INF);fill(vis,vis+maxn,false);d[s]=0;c[s]=0;for(int i=0;i<n;i++) pre[i]=i;for(int i=0;i<n;i++){int u=-1,MIN=INF;for(int j=0;j<n;j++){if(vis[j]==false&&d[j]<MIN){MIN=d[j];u=j;}}if(u==-1) return;vis[u]=true;for(int v=0;v<n;v++){if(vis[v]==false&&G[u][v]!=INF){if(d[u]+G[u][v]<d[v]){d[v]=d[u]+G[u][v];c[v]=cost[u][v]+c[u];pre[v]=u;}else if(d[u]+G[u][v]==d[v]){if(c[v]>c[u]+cost[u][v]){c[v]=cost[u][v]+c[u];pre[v]=u;}}}}}
}
void dfs(int ed){if(ed==st){printf("%d ",ed);return;} dfs(pre[ed]);printf("%d ",ed);
}
int main(){scanf("%d%d%d%d",&n,&m,&st,&ed);fill(G[0],G[0]+maxn*maxn,INF);while(m--){int v1,v2;scanf("%d%d",&v1,&v2);scanf("%d%d",&G[v1][v2],&cost[v1][v2]);G[v2][v1]=G[v1][v2];cost[v2][v1]=cost[v1][v2];}dijkst(st);dfs(ed);printf("%d %d",d[ed],c[ed]);return 0;
}

A1031

#include<cstdio>
#include<cstring>
int main() {char str[100], uu[40][40];scanf("%s",str); int N = strlen(str);int n1=(N+2)/3;int n3=(N+2)/3;int n2=N-n1-n3+2;int pos=0;memset(uu,' ',sizeof(uu));for(int i=0;i<n1-1;i++){uu[i][0]=str[pos++];} for(int j=0;j<n2;j++){uu[n1-1][j]=str[pos++];}for(int k=n3-2;k>=0;k--){uu[k][n2-1]=str[pos++];}for(int i=0;i<n1;i++){for(int j=0;j<n2;j++){printf("%c",uu[i][j]);}printf("\n");}return 0;
}

A1032(25 链表 找公共后缀)

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100010;
struct node{char data;bool flag;int next;
}list[maxn];
int main(){int address,begin1,begin2,n;scanf("%d%d%d",&begin1,&begin2,&n);for(int i=0;i<n;i++){scanf("%d",&address);scanf(" %c %d",&list[address].data,&list[address].next);}while(begin1!=-1){list[begin1].flag=true;begin1=list[begin1].next;}while(begin2!=-1){if(list[begin2].flag==true) {printf("%05d",begin2);return 0;}begin2=list[begin2].next;}printf("-1");return 0;
}

A1033

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 510;
const int INF = 1000000000;
struct node{double price;double distance;
}station[maxn];
bool cmp(node a, node b) {return a.distance < b.distance;
}
int main() {double cmax, d, davg;int n;scanf("%lf%lf%lf%d", &cmax, &d, &davg, &n);for (int i = 0; i < n; i++) {scanf("%lf%lf", &station[i].price, &station[i].distance);}station[n].distance = d;station[n].price = 0;sort(station, station + n + 1, cmp);if (station[0].distance != 0) {printf("The maximum travel distance = 0.00");return 0;}int now = 0;double nowtank = 0, expend = 0,  canrun = cmax * davg;while (now < n) {int k=-1;double minprice=INF;for (int i = now + 1; (station[i].distance - station[now].distance) <= canrun && i <= n; i++) {if (station[i].price < minprice) {minprice = station[i].price;k = i;if (minprice < station[now].price) {break;}}}if (k == -1) break;double need = (station[k].distance - station[now].distance) / davg;if (station[k].price < station[now].price) {if (nowtank < need) {expend += (need - nowtank) * station[now].price;nowtank = 0;}else {nowtank -= need;}}else {expend += (cmax - nowtank) * station[now].price;nowtank = cmax - need;}now = k;}if (now == n) printf("%.2f", expend);else printf("The maximum travel distance = %.2f", station[now].distance + canrun);return 0;
}

A1034(30 图的深度遍历、map、连通块、边权和点权)

#include<iostream>
#include<map>
#include<string>
using namespace std;
map<string,int> strtoint;
map<int,string> inttostr;
map<string,int> ans;
bool vis[2020];
int G[2020][2020],weight[2020],n,k,index=1;
int Stringtoint(string s){if(strtoint.count(s)==0){strtoint[s]=index;inttostr[index]=s;index++;//同时计数总的结点个数 }return strtoint[s];
}
void dfs(int id,int &totaltime,int&head,int &num){vis[id]=true;num++;if(weight[id]>weight[head]) head=id;for(int i=1;i<index;i++){if(G[id][i]!=0){totaltime+=G[id][i];G[id][i]=G[i][id]=0;if(vis[i]==false) dfs(i,totaltime,head,num);} }
}
int main(){string s1,s2;cin>>n>>k;int id1,id2,calltime;for(int i=0;i<n;i++){cin>>s1>>s2>>calltime;id1=Stringtoint(s1),id2=Stringtoint(s2);G[id1][id2]+=calltime;G[id2][id1]+=calltime;weight[id1]+=calltime;weight[id2]+=calltime;}fill(vis,vis+2020,false);for(int i=1;i<index;i++){if(vis[i]==false){int totaltime=0,head=i,num=0;dfs(i,totaltime,head,num);if(num>2&&totaltime>k) ans[inttostr[head]]=num;}}cout<<ans.size()<<endl;for(auto it=ans.begin();it!=ans.end();it++){cout<<it->first<<" "<<it->second<<endl;}return 0;
}

A1035(20 字符串)

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){int n;cin>>n;vector<string> v;for(int j=0;j<n;j++){bool flag=false;string name,password,ans;cin>>name>>password;for(int i=0;i<password.length();i++){switch(password[i]){case '1':password[i]='@';flag=true;break;case '0':password[i]='%';flag=true;break;case 'l':password[i]='L';flag=true;break;case 'O':password[i]='o';flag=true;break;}}if(flag==true) v.push_back(name+" "+password);}if(n==1) printf("There is 1 account and no account is modified\n");else if(v.size()==0) printf("There are %d accounts and no account is modified\n",n);else{printf("%d\n",v.size());for(int i=0;i<v.size();i++) cout<<v[i]<<endl;}return 0;
}

A1036

#include<cstdio>
#include<algorithm>
using namespace std;
struct student {char name[11];char gender;char id[11];int score;
}male,female,temp;
int main() {int n;scanf("%d", &n);female.score = -1;male.score = 101;for (int i = 0; i < n; i++) {scanf("%s %c %s %d", temp.name, &temp.gender, temp.id, &temp.score);if (temp.gender == 'F') {if (temp.score > female.score) {female = temp;}}if (temp.gender == 'M') {if (temp.score < male.score) {male = temp;}}}if (female.score == -1) printf("Absent\n");if (female.score > -1) printf("%s %s\n", female.name, female.id);if (male.score == 101) printf("Absent\n");if (male.score <101) printf("%s %s\n", male.name, male.id);if (female.score == -1 || male.score == 101) printf("NA");if (female.score > -1 && male.score < 101) printf("%d", female.score - male.score);return 0;
}

A1037(25 贪心)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
bool cmp(int a,int b){return a>b;}
int main(){int nc,np,c,p,ans=0;vector<int>c1,c2,p1,p2;scanf("%d",&nc);for(int i=0;i<nc;i++){scanf("%d",&c);if(c>=0) c1.push_back(c);else c2.push_back(c);}scanf("%d",&np);for(int i=0;i<np;i++){scanf("%d",&p);if(p>=0) p1.push_back(p);else p2.push_back(p);}sort(c1.begin(),c1.end(),cmp);sort(p1.begin(),p1.end(),cmp);sort(c2.begin(),c2.end());sort(p2.begin(),p2.end());for(int i=0;i<c1.size()&&i<p1.size();i++) ans+=c1[i]*p1[i];for(int i=0;i<c2.size()&&i<p2.size();i++)   ans+=c2[i]*p2[i];printf("%d",ans);return 0;
}

A1038(30 贪心 求最小字符串序列)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
bool cmp(string &a,string &b){return a+b<b+a;
}
int main(){int n;cin>>n;string ans;vector<string> v(n);for(int i=0;i<n;i++) cin>>v[i];sort(v.begin(),v.end(),cmp);for(int i=0;i<n;i++) ans+=v[i];while(ans[0]=='0') ans.erase(0,1);if(ans.size()==0) cout<<"0";else cout<<ans;return 0;
}

A1039(25 vector)

#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=26*26*26*10+10;
int getid(char *name){int id=0;for(int i=0;i<3;i++) id=26*id+(name[i]-'A');id=id*10+(name[3]-'0');return id;
}
int main(){int n,k,courseid,num;char name[4];scanf("%d%d",&n,&k);vector<int> v[maxn];for(int i=0;i<k;i++){scanf("%d%d",&courseid,&num);for(int j=0;j<num;j++){scanf("%s",name);v[getid(name)].push_back(courseid);}}for(int i=0;i<n;i++){scanf("%s",name);printf("%s %d",name,v[getid(name)].size());sort(v[getid(name)].begin(),v[getid(name)].end());for(int j=0;j<v[getid(name)].size();j++) printf(" %d",v[getid(name)][j]);printf("\n");}return 0;
}

A1040

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
const int maxn=1010;
string str;
int dp[maxn][maxn];
int main(){memset(dp,0,sizeof(dp));getline(cin,str);int ans=1,len=str.length();for(int i=0;i<len;i++){dp[i][i]=1;if(i<len-1){if(str[i]==str[i+1]){dp[i][i+1]=1;ans=2;}}}for(int L=3;L<=len;L++){for(int i=0;i+L-1<len;i++){int j=i+L-1;if(str[i]==str[j]&&dp[i+1][j-1]==1){dp[i][j]=1;ans=L;}}}printf("%d",ans);return 0;
}

A1041(20 hash)

#include<string>
#include<vector>
#include<iostream>
using namespace std;
int main(){int n,flag[100001]={0},j;cin>>n;vector<int> v(n);for(int i=0;i<n;i++){scanf("%d",&v[i]);flag[v[i]]++;}for(j=0;j<n;j++){if(flag[v[j]]==1) {cout<<v[j];break;}}if(j==n) cout<<"None";return 0;
}

A1042

#include<cstdio>
const int N = 55;
int main()
{char mp[5] = { 'S','H','C','D','J' };int start[N], next[N], end[N];int times;scanf("%d", &times);for (int i = 1; i <= 54; i++) {scanf("%d", &next[i]);}for (int i = 1; i <= 54; i++) {start[i] = i;}for (int step = 0; step < times; step++) {for (int i = 1; i <= 54; i++) {end[next[i]] = start[i];}for (int i = 1; i <= 54; i++) {start[i] = end[i];}}for (int i = 1; i <= 54; i++) {if (i != 1) printf(" ");start[i]--;printf("%c%d", mp[start[i] / 13], start[i] % 13+1);}return 0;
}

A1043(25 二叉搜索树的遍历与反转)

#include<cstdio>
#include<vector>
using namespace std;
struct node{int data;node*left,*right;
};
void insert(node*&root,int data){if(root==NULL){root=new node;root->data=data;root->left=NULL;root->right=NULL;return;}if(root->data<=data) insert(root->right,data);else if(root->data>data) insert(root->left,data);
}
void preorder(node*root,vector<int>&v){if(root==NULL) return;v.push_back(root->data);preorder(root->left,v);preorder(root->right,v);
}
void preorderM(node*root,vector<int>&v){if(root==NULL) return;v.push_back(root->data);preorderM(root->right,v);preorderM(root->left,v);
}
void postorder(node*root,vector<int>&v){if(root==NULL) return;postorder(root->left,v);postorder(root->right,v);v.push_back(root->data);
}
void postorderM(node*root,vector<int>&v){if(root==NULL) return;postorderM(root->right,v);postorderM(root->left,v);v.push_back(root->data);
}
int main(){int n;scanf("%d",&n);vector<int> v(n),pre,preM,post,postM;node*root=NULL;for(int i=0;i<n;i++) {scanf("%d",&v[i]);insert(root,v[i]);}preorder(root,pre);preorderM(root,preM);if(pre==v){printf("YES\n");postorder(root,post);for(int i=0;i<n;i++){if(i>0) printf(" ");printf("%d",post[i]);}}else if(preM==v){printf("YES\n");postorderM(root,postM);for(int i=0;i<n;i++){if(i>0) printf(" ");printf("%d",postM[i]);}}else printf("NO");return 0;
}

A1044(25 二分)

#include<iostream>
#include<vector>
using namespace std;
vector<int> sum,ans;
int n,m;
void func(int i,int &j,int &tempsum){int low=i,high=n;while(low<high){int mid=(low+high)/2;if(sum[mid]-sum[i-1]>=m) high=mid;else low=mid+1;}j=high;tempsum=sum[j]-sum[i-1];
}
int main(){cin>>n>>m;sum.resize(n+1);for(int i=1;i<=n;i++){cin>>sum[i];sum[i]+=sum[i-1];}int minans=sum[n];for(int i=1;i<=n;i++){int j,tempsum;func(i,j,tempsum);if(tempsum>minans) continue;if(tempsum>=m){if(tempsum<minans){ans.clear();minans=tempsum;}ans.push_back(i);ans.push_back(j);}}for(int i=0;i<ans.size();i+=2)  printf("%d-%d\n",ans[i],ans[i+1]);return 0;
}

A1045

方法一:最长不下降子序列LIS
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=10010;
int hashtable[maxn],a[maxn],dp[maxn];
int main(){int num=0,ans=-1,n,m,L,temp;scanf("%d",&n);scanf("%d",&m);fill(hashtable,hashtable+maxn,-1);for(int i=0;i<m;i++){scanf("%d",&temp);hashtable[temp]=i;}scanf("%d",&L);for(int i=0;i<L;i++){scanf("%d",&temp);if(hashtable[temp]>=0){a[num++]=hashtable[temp];}}for(int i=0;i<num;i++){dp[i]=1;for(int j=0;j<i;j++){if(a[i]>=a[j]&&dp[j]+1>dp[i]){dp[i]=dp[j]+1;}}ans=max(ans,dp[i]);}printf("%d",ans);return 0;
}
方法二:最长公共子序列
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=10010;
int a[maxn],b[maxn],dp[maxn][maxn];
int main(){int n,m,L;scanf("%d",&n);scanf("%d",&m);for(int i=1;i<=m;i++) scanf("%d",&a[i]);scanf("%d",&L);for(int i=1;i<=L;i++) scanf("%d",&b[i]);for(int i=0;i<m;i++) dp[i][0]=0;for(int i=0;i<L;i++) dp[0][i]=0;for(int i=1;i<=m;i++){for(int j=1;j<=L;j++){int maxone=max(dp[i-1][j],dp[i][j-1]);if(a[i]==b[j]) dp[i][j]=maxone+1;else dp[i][j]=maxone;}}printf("%d",dp[m][L]);return 0;
}

A1046

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int dis[maxn], A[maxn];
int main() {int n,times,left,right,sum=0;scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &A[i]);sum += A[i];dis[i] = sum;}scanf("%d", &times);for (int i = 0; i < times; i++) {scanf("%d%d", &left, &right);if (left > right)  swap(left, right);int distance = dis[right - 1] - dis[left - 1];printf("%d\n", min(distance,sum-distance));}return 0;
}

A1047(25 vector)

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
char name[40010][5];
vector<int> v[2510];
bool cmp(int a,int b){return strcmp(name[a],name[b])<0;
}
int main(){int n,k,c,courseid;scanf("%d%d",&n,&k);for(int i=0;i<n;i++){scanf("%s%d",name[i],&c);while(c--){scanf("%d",&courseid);v[courseid].push_back(i);}}for(int i=1;i<=k;i++){printf("%d %d\n",i,v[i].size());sort(v[i].begin(),v[i].end(),cmp);for(int j=0;j<v[i].size();j++) printf("%s\n",name[v[i][j]]);}return 0;
}

A1048(25 )

1.哈希散列解法
#include<cstdio>
int main(){int n,pay,flag[100010]={0},a[100010];scanf("%d%d",&n,&pay);for(int i=0;i<n;i++) {scanf("%d",&a[i]);flag[a[i]]++;}for(int i=1;i<pay;i++){if(flag[i]>0&&flag[pay-i]>0){if((i==pay-i&&flag[i]>1)||i!=pay-i){printf("%d %d",i,pay-i);return 0;}}} printf("No Solution");return 0;
}2.二分法解法
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int main() {int n, m;int num[maxn];scanf("%d%d", &n, &m);for (int i = 0; i < n; i++) {scanf("%d", &num[i]);}sort(num, num + n);int j,i;for ( i = 0; i < n; i++) {j = lower_bound(num + i, num +n, m - num[i]) - num;if ( (num[j] + num[i] == m&&j!=i)||(num[i]==num[i+1]&&num[i]*2==m)) {printf("%d %d", num[i], num[j]);break;}}if (i >= n) printf("No Solution");return 0;
}3.two points
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int main() {int n, m;int a[maxn];scanf("%d%d", &n, &m);for (int i = 0; i < n; i++) {scanf("%d", &a[i]);}sort(a, a + n);int j=n-1,i=0;while(i<j){if(a[i]+a[j]==m){printf("%d %d",a[i],a[j]);return 0;}else if(a[i]+a[j]>m) j--;else i++;}printf("No Solution");return 0;
}

A1049(30 数学问题："1"的个数)

#include<iostream>
using namespace std;
int main(){int n,left=0,right=0,a=1,now=1,ans=0;cin>>n;while(n/a){left=n/(a*10),right=n%a,now=n/a%10;if(now==0) ans+=left*a;else if(now==1) ans+=left*a+right+1;else ans+=(left+1)*a;a*=10; }cout<<ans;return 0;
}

A1050(20 hash)

#include<string>
#include<iostream>
#include<vector>
using namespace std;
int main(){string s1,s2,ans;vector<int> v(128,0);getline(cin,s1);getline(cin,s2);for(int i=0;i<s2.length();i++) v[s2[i]]++;for(int i=0;i<s1.length();i++){if(v[s1[i]]==0) ans+=s1[i];}cout<<ans;return 0;
}

A1051(25 栈)

#include<cstdio>
#include<stack>
#include<vector>
using namespace std;
int main(){int n,m,k;scanf("%d%d%d",&m,&n,&k);while(k--){vector<int> v(n+1);stack<int> st;bool flag=true;int current=1;for(int i=1;i<=n;i++) scanf("%d",&v[i]);for(int i=1;i<=n;i++){st.push(i);if(st.size()>m){flag=false;break;}while(!st.empty()&&st.top()==v[current]){st.pop();current++;}}if(current!=n+1) flag=false;if(flag) printf("YES\n");else printf("NO\n");}return 0;
}

A1052(25 链表按key排序)

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=100010;
struct node{int data,address,next;
}list[maxn];
bool cmp(node a,node b){return a.data<b.data;}
int main(){int n,begin,address;vector<node> v;scanf("%d%d",&n,&begin);for(int i=0;i<n;i++){scanf("%d",&address);scanf("%d%d",&list[address].data,&list[address].next);list[address].address=address;}while(begin!=-1){v.push_back(list[begin]);begin=list[begin].next;}if(v.size()==0) printf("0 -1");else{sort(v.begin(),v.end(),cmp);printf("%d %05d\n",v.size(),v[0].address);for(int i=0;i<v.size()-1;i++) printf("%05d %d %05d\n",v[i].address,v[i].data,v[i+1].address);printf("%05d %d -1\n",v[v.size()-1].address,v[v.size()-1].data);}return 0;
}

A1053(30 输出权值总和同为给定值的路径(dfs))

1.#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n,m,s,sum=0,path[101];
struct node{int w;vector<int> child;
}tree[101];
bool cmp(int a,int b){return tree[a].w>tree[b].w;}
void dfs(int index,int nodenum,int sum){if(sum>s) return;if(sum==s){if(tree[index].child.size()!=0) return;for(int i=0;i<nodenum;i++){if(i>0) printf(" ");printf("%d",tree[path[i]].w);}printf("\n");return;}for(int i=0;i<tree[index].child.size();i++){int child=tree[index].child[i];path[nodenum]=child;dfs(child,nodenum+1,sum+tree[child].w);}
}
int main(){int id,k,c;scanf("%d%d%d",&n,&m,&s);for(int i=0;i<n;i++) scanf("%d",&tree[i].w);for(int i=0;i<m;i++){scanf("%d%d",&id,&k);for(int j=0;j<k;j++){scanf("%d",&c);tree[id].child.push_back(c);}sort(tree[id].child.begin(),tree[id].child.end(),cmp);}path[0]=0;dfs(0,1,tree[0].w);return 0;
}2.
#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
int n,m,s,weight[maxn],notroot[maxn]={0};
vector<int> tree[maxn];
vector<int> t,path;
bool cmp(int a,int b){return weight[a]>weight[b];
}
void dfs(int root,int sum){if(tree[root].size()!=0&&sum>s)  return;if(tree[root].size()==0&&sum==s){t.push_back(root);path=t;for(int i=0;i<path.size();i++){if(i>0) printf(" ");printf("%d",weight[path[i]]);}printf("\n");path.clear();t.pop_back();return;}t.push_back(root);for(int i=0;i<tree[root].size();i++) dfs(tree[root][i],sum+weight[tree[root][i]]);t.pop_back();
}
int main(){int root=0;cin>>n>>m>>s;for(int i=0;i<n;i++) cin>>weight[i];for(int i=0;i<m;i++){int id,k,temp;cin>>id>>k;while(k--){cin>>temp;notroot[temp]=1;tree[id].push_back(temp);}sort(tree[id].begin(),tree[id].end(),cmp);} while(notroot[root]!=0&&root<n) root++;dfs(root,weight[root]);return 0;
}

A1054(20 map)

#include<cstdio>
#include<map>
using namespace std;
int main(){int m,n,temp;scanf("%d%d",&m,&n);map<int,int> mp;for(int i=0;i<n;i++){for(int j=0;j<m;j++){scanf("%d",&temp);mp[temp]++;}}int times=-1,ans;for(auto it=mp.begin();it!=mp.end();it++){if(it->second>times){times=it->second;ans=it->first;} } printf("%d",ans);return 0;
}

A1055(25 按条件查找并排序输出)

#include<algorithm>
#include<vector>
#include<cstring>
#include<iostream>
using namespace std;
struct node{char name[10];int age,worth;
};
bool cmp(node a,node b){if(a.worth!=b.worth) return a.worth>b.worth;else if(a.age!=b.age) return a.age<b.age;else return strcmp(a.name,b.name)<0;
}
int main(){int n,m,k,amin,amax;scanf("%d%d",&n,&m);vector<node> t(n),v;vector<int> book(201,0);for(int i=0;i<n;i++)  scanf("%s%d%d",t[i].name,&t[i].age,&t[i].worth);sort(t.begin(),t.end(),cmp);   for(int i=0;i<n;i++){if(book[t[i].age]<100){v.push_back(t[i]);book[t[i].age]++;}}for(int i=1;i<=m;i++) {printf("Case #%d:\n",i);vector<node> ans;scanf("%d%d%d",&k,&amin,&amax);for(int j=0;j<v.size();j++) {if(v[j].age>=amin&&v[j].age<=amax) ans.push_back(v[j]);}if(ans.size()==0) printf("None\n");else if(k>=ans.size()){for(int j=0;j<ans.size();j++) printf("%s %d %d\n",ans[j].name,ans[j].age,ans[j].worth);}else if(k<ans.size())  {for(int j=0;j<k;j++) printf("%s %d %d\n",ans[j].name,ans[j].age,ans[j].worth);}   }return 0;
}

A1056(25 queue队列)

#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
struct node{int w,rank;
};
int main(){int np,ng,orderid;queue<int> q;scanf("%d%d",&np,&ng);vector<node> v(np);for(int i=0;i<np;i++) scanf("%d",&v[i].w);for(int i=0;i<np;i++) {scanf("%d",&orderid);q.push(orderid);}int temp=np,group;while(q.size()!=1){if(temp%ng==0) group=temp/ng;else group=temp/ng+1;for(int i=0;i<group;i++){int k=q.front();for(int j=1;j<=ng;j++){if(i*ng+j>temp) break;int front=q.front();if(v[front].w>v[k].w) k=front;v[front].rank=group+1;q.pop();}q.push(k); }temp=group; }v[q.front()].rank=1;for(int i=0;i<np;i++){if(i!=0) printf(" ");printf("%d",v[i].rank);}return 0;
}

A1057(30 分块、树状数组)

1.分块
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int table[100010],block[317];
stack<int> st;
void Peekmedian(int x) {int sum = 0, block_id = 0;while (sum + block[block_id] < x) {sum += block[block_id++];}int num = block_id * 316;while (sum + table[num] < x) sum+=table[num++];printf("%d\n", num);
}
void Push(int x) {st.push(x);table[x]++;block[x / 316]++;
}
void Pop() {int top = st.top();st.pop();table[top]--;block[top / 316]--;printf("%d\n", top);
}
int main() {int n, temp;char cmd[15];scanf("%d", &n);memset(table, 0, sizeof(table));memset(block, 0, sizeof(block));for (int i = 0; i < n; i++) {scanf("%s", cmd);if (strcmp(cmd, "Pop")==0) {if (st.empty()) printf("Invalid\n");else Pop();}else if (strcmp(cmd, "Push")==0) {scanf("%d", &temp);Push(temp);}else {if (st.empty()) printf("Invalid\n");else {int k = st.size();if (k % 2 == 0) k = k / 2;else k = k / 2 + 1;Peekmedian(k);}}}return 0;
}
2.树状数组

A1058(20 水题)

#include<cstdio>
int main(){int a,b,c,d,e,f,ans[3];scanf("%d.%d.%d %d.%d.%d",&a,&b,&c,&d,&e,&f);ans[2]=(c+f)%29;ans[1]=(b+e+(c+f)/29)%17;ans[0]=a+d+(b+e+(c+f)/29)/17;for(int i=0;i<3;i++){if(i!=0) printf(".");printf("%d",ans[i]);}return 0;
}

A1059(25 建立素数表 求素数因数)

#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int maxn=105000;
vector<int> prime;
long long n;
bool isprime(int n) {if(n<=1) return false;int sqr=sqrt(n*1.0);for(int i=2; i<=sqr; i++) {if(n%i==0) return false;}return true;
}
int main() {int index=0;scanf("%lld",&n);printf("%lld=",n);for(int i=2; i<=maxn; i++) {if(isprime(i)) prime.push_back(i);}if(n==1) printf("1");else {int hasprint=0;for(int i=0; n>=2; i++) {int cnt=0,flag=0;while(n%prime[i]==0) {n/=prime[i];cnt++;flag=1;}if(flag){if(hasprint) printf("*");printf("%lld",prime[i]);hasprint=1;}if(cnt>=2) printf("^%d",cnt);}}return 0;
}

A1060(25 字符串 科学计数法)

#include<iostream>
#include<string>
using namespace std;
int n;
string deal(string s,int &e){int k=0;while(s[0]=='0'&&s.length()>0) s.erase(s.begin());//去除前导0 if(s[0]=='.'){//若去除后是小数点，说明是小于1的数 s.erase(s.begin());while(s[0]=='0'&&s.length()>0) {s.erase(s.begin());//去除小数点后非零位前的所有0 e--;}}else{while(s[k]!='.'&&k<s.length()){//寻找小数点 k++;e++;}if(k<s.length()) s.erase(s.begin()+k);//删除小数点 }if(s.length()==0) e=0;int num=0;k=0;string ans;while(num<n){if(k<s.length()) ans+=s[k++];else ans+='0';num++;}   return ans;
}
int main(){string s1,s2,s3,s4;cin>>n>>s1>>s2;int e1=0,e2=0;s3=deal(s1,e1);s4=deal(s2,e2);if(s3==s4&&e1==e2) cout<<"YES"<<" 0."<<s3<<"*10^"<<e1;   else cout<<"NO"<<" 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2;return 0;
}

A1061(20 字符串)

#include<iostream>
#include<string>
#include<cctype>
using namespace std;
int main(){int i=0,pos=0;char ans[2];string a,b,c,d;cin>>a>>b>>c>>d;while(i<a.length()&&i<b.length()){if(a[i]==b[i]&&(a[i]>='A'&&a[i]<='G')){ans[0]=a[i];break;}i++;}++i;while(i<a.length()&&i<b.length()){if(a[i]==b[i]&&((a[i]>='A'&&a[i]<='N')||isdigit(a[i]))){ans[1]=a[i];break;}i++;}while(pos<c.length()&&pos<d.length()){if(c[pos]==d[pos]&&isalpha(c[pos])) break;pos++;}string weekday[7]={"MON","TUE","WED","THU","FRI","SAT","SUN"};printf("%s %02d:%02d",weekday[ans[0]-'A'].c_str(),isdigit(ans[1])?ans[1]-'0':ans[1]-'A'+10,pos);return 0;
}

A1062

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
struct node{int num;int de,cai,rank;
}temp;
bool cmp( node a,node b){if(a.rank!=b.rank) return a.rank<b.rank;else if((a.de+a.cai)!=(b.de+b.cai)) return (a.de+a.cai)>(b.de+b.cai);else if(a.de!=b.de) return a.de>b.de;else return a.num<b.num;
}
int main(){int n,L,H;scanf("%d%d%d",&n,&L,&H);vector<node> v;for(int i=0;i<n;i++){int id,a,b,rank;scanf("%d%d%d",&id,&a,&b);if(a<L||b<L) continue;if(a>=H&&b>=H) rank=1;else if(a>=H&&b<H) rank=2;else if(a<H&&b<H&&a>=b) rank=3;else rank=4;v.push_back(node{id,a,b,rank});}sort(v.begin(),v.end(),cmp);printf("%d\n",v.size());for(int i=0;i<v.size();i++) printf("%08d %d %d\n",v[i].num,v[i].de,v[i].cai);return 0;
}

A1063(25 set)

#include<cstdio>
#include<set>
using namespace std;
const int maxn = 51;
set<int> v[maxn];
int main() {int n,m,temp,k;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &m);for (int j = 0; j < m; j++) {scanf("%d", &temp);v[i].insert(temp);}}int a, b;scanf("%d", &k);for (int i = 0; i < k; i++) {scanf("%d%d", &a, &b);int nc=0, nt=v[b-1].size();for (auto it = v[a - 1].begin(); it != v[a - 1].end(); it++) {if (v[b - 1].find(*it) != v[b - 1].end())  nc++;else nt++;}printf("%.1f%%\n", (nc * 100.0) / nt);}
}

A1064(30 完全二叉树(中序遍历建树))

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1001;
int CBT[maxn],a[maxn],n,index=0;
void inorder(int root){if(root>n) return;inorder(root*2);CBT[root]=a[index++];inorder(root*2+1);
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&a[i]);sort(a,a+n);inorder(1);for(int i=1;i<=n;i++){if(i>1) printf(" ");printf("%d",CBT[i]);}return 0;
}

A1065

#include<cstdio>
int main() {long long a, b, c;int n,times=1;scanf("%d", &n);while (n--) {scanf("%lld%lld%lld", &a, &b, &c);bool flag;long long sum = a + b;if (a > 0 && b > 0 && sum < 0)flag = true;else if (a < 0 && b < 0 && sum >= 0)flag = false;else if (sum > c)flag = true;else flag = false;if (flag == true)printf("Case #%d:true", times++);if (flag == false)printf("Case #%d:false", times++);}return 0;
}

A1066(25 建立平衡二叉树)

#include<cstdio>
#include<algorithm>
using namespace std;
struct node{int v,height;node*left,*right;
};
node* newnode(int v){node* Node=new node;Node->v=v;Node->height=1;Node->left=Node->right=NULL;return Node;
}
int getheight(node*root){if(root==NULL) return 0;return root->height;
}
void updateheight(node *root){root->height=max(getheight(root->right),getheight(root->left))+1;
}
int getbalance(node* root){return getheight(root->left)-getheight(root->right);
}
void L(node*&root){node*temp=root->right;root->right=temp->left;temp->left=root;updateheight(root);updateheight(temp);root=temp;
}
void R(node*&root){node*temp=root->left;root->left=temp->right;temp->right=root;updateheight(root);updateheight(temp);root=temp;
}
void insert(node*&root,int v){if(root==NULL){root=newnode(v);return;}if(v<root->v){insert(root->left,v);updateheight(root);if(getbalance(root)==2){if(getbalance(root->left)==1) R(root);else if(getbalance(root->left)==-1){L(root->left);R(root);}}}else{insert(root->right,v);updateheight(root);if(getbalance(root)==-2){if(getbalance(root->right)==-1) L(root);else if(getbalance(root->right)==1){R(root->right);L(root);}}}
}
int main(){int n,temp;node* root=NULL;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&temp);insert(root,temp);}printf("%d",root->v);return 0;
}

A1067(25 贪心 swap排序，求最小交换次数)

#include<iostream>
#include<algorithm>
using namespace std;
int main(){int n,t,cnt=0,a[100001];cin>>n;for(int i=0;i<n;i++){cin>>t;a[t]=i;}for(int i=1;i<n;i++){while(a[0]!=0){swap(a[0],a[a[0]]);cnt++;}if(a[i]!=i){swap(a[i],a[0]);cnt++;}} cout<<cnt;return 0;
}

A1068

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=10010;
const int maxv=110;
int dp[maxv]={0},w[maxn];
bool choice[maxn][maxv],flag[maxn];
bool cmp(int a,int b){return a>b;
}
int main(){int n,m;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&w[i]);}sort(w+1,w+1+n,cmp);for(int i=1;i<=n;i++){for(int v=m;v>=w[i];v--){if(dp[v]<=dp[v-w[i]]+w[i]){dp[v]=dp[v-w[i]]+w[i];choice[i][v]=1;}else{choice[i][v]=0;}}}if(dp[m]!=m) printf("No Solution");else{int k=n,num=0,v=m;while(k>=0){if(choice[k][v]==1){flag[k]=true;v-=w[k];num++;}else flag[k]=false;k--;}for(int i=n;i>0;i--){if(flag[i]==true){printf("%d",w[i]);num--;if(num>0) printf(" ");}}}return 0;
}

A1069(20 数字黑洞)

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string strsort(string s){sort(s.begin(),s.end());s.insert(0,4-s.size(),'0');return s;
}
string rev(string s){reverse(s.begin(),s.end());return s;
}
int main(){string s;cin>>s;s.insert(0, 4 - s.length(), '0');if(rev(s)==s){cout<<s<<" - "<<s<<" = "<<"0000";return 0;}int ans=1;while(1){s=strsort(s);string s1=rev(s);ans=stoi(s1)-stoi(s);printf("%s - %s = %04d\n",s1.c_str(),s.c_str(),ans);if(ans==6174) return 0;s=to_string(ans);}return 0;
}

A1070(25 贪心)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{double inventory,price;
};
bool cmp(node &a,node &b){ return a.price>b.price;};
int main(){int n,D;scanf("%d%d",&n,&D);double prices,profit=0.0;vector<node> v(n);for(int i=0;i<n;i++) scanf("%lf",&v[i].inventory);for(int i=0;i<n;i++){scanf("%lf",&prices);v[i].price=prices/v[i].inventory;}sort(v.begin(),v.end(),cmp);for(int i=0;i<n;i++){if(v[i].inventory>=D){profit+=D*v[i].price;break;}else{profit+=v[i].inventory*v[i].price;D-=v[i].inventory;}}printf("%.2f",profit);return 0;
}

A1071(25 map 字符串)

#include<string>
#include<map>
#include<cctype>
#include<iostream>
using namespace std;
int main(){map<string,int> mp;string s,temp,ans;getline(cin,s);int maxtimes=-1;for(int i=0;i<s.length();i++){if(isalnum(s[i])){if(isupper(s[i])) s[i]=tolower(s[i]);temp+=s[i];}if(!isalnum(s[i])||i==s.length()-1){if(!temp.empty()) mp[temp]++;temp.clear();}}for(auto it=mp.begin();it!=mp.end();it++){if(it->second>maxtimes) {ans=it->first;maxtimes=it->second;}}cout<<ans<<" "<<maxtimes;return 0;
}

A1072(30 多源dijkst求最佳源)

#include<iostream>
#include<string>
#include<algorithm>
#include<map>
using namespace std;
const int maxn=1100;
const int INF=0x7fffffff;
int n,m,k,ds,G[maxn][maxn],d[maxn],index_g=n+1;
bool vis[maxn];
map<string,int> mp;
void dijkst(int s){fill(d,d+maxn,INF);d[s]=0;fill(vis,vis+maxn,false);for(int i=1;i<=n+m;i++){int u=-1,MIN=INF;for(int j=1;j<=n+m;j++){if(vis[j]==false&&d[j]<MIN){u=j;MIN=d[j];}}if(u==-1) return;vis[u]=true;for(int v=1;v<=n+m;v++){if(vis[v]==false&&G[u][v]!=INF){if(d[u]+G[u][v]<d[v]) d[v]=d[u]+G[u][v];}}}
}
int getid(string s){if(s[0]=='G'){s.erase(s.begin());if(mp.count(s)==0) mp[s]=stoi(s)+n;return mp[s];}else return stoi(s);
}
int main(){int ansid=-1;double ansavg=INF,ansmindis=-1;string s1,s2;fill(G[0],G[0]+maxn*maxn,INF);scanf("%d%d%d%d",&n,&m,&k,&ds);while(k--){cin>>s1>>s2;int v1=getid(s1);int v2=getid(s2);cin>>G[v1][v2];G[v2][v1]=G[v1][v2];}for(int i=n+1;i<=n+m;i++){double mindis=INF,avg=0,sum=0;dijkst(i);for(int j=1;j<=n;j++){if(d[j]>ds){mindis=-1;break;}if(d[j]<mindis) mindis=d[j];avg+=1.0*d[j]/n;}if(mindis==-1) continue;if(mindis>ansmindis){ansid=i;ansmindis=mindis;ansavg=avg;}else if(mindis==ansmindis&&ansavg>avg){ansid=i;ansavg=avg;}}if(ansid==-1) printf("No Solution");else{printf("G%d\n",ansid-n);printf("%.1f %.1f",ansmindis,ansavg);}return 0;
}

A1073 科学计数法

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main(){string s,t;int n,i=0;cin>>s;while(s[i]!='E') i++;t=s.substr(1,i-1);n=stoi(s.substr(i+1,s.length()-i-1));if(s[0]=='-') cout<<"-";if(n<0){cout<<"0.";for(int j=0;j<abs(n)-1;j++) cout<<"0";for(int j=0;j<t.size();j++) if(t[j]!='.') cout<<t[j];}else{cout<<t[0];int cnt,j;for(j=2,cnt=0;j<t.length()&&cnt<n;j++,cnt++) cout<<t[j];if(j==t.length()) {for(int k=0;k<n-cnt;k++) cout<<"0";} else{cout<<".";for(int k=j;k<t.size();k++) cout<<t[k];}}return 0;
}

A1074(25 反转链表)

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100010;
struct node{int address,data,next;int order;
}list[maxn];
bool cmp(node &a,node &b){return a.order<b.order;
}
int main(){int begin,n,k,address,cnt=0;for(int i=0;i<maxn;i++想·) list[i].order=maxn;scanf("%d%d%d",&begin,&n,&k);for(int i=0;i<n;i++){scanf("%d",&address);scanf("%d%d",&list[address].data,&list[address].next);list[address].address=address;}int p=begin;while(p!=-1){list[p].order=cnt++;p=list[p].next;}sort(list,list+maxn,cmp);for(int i=0;i<cnt/k;i++){for(int j=(i+1)*k-1;j>i*k;j--){printf("%05d %d %05d\n",list[j].address,list[j].data,list[j-1].address);}printf("%05d %d ",list[i*k].address,list[i*k].data);if(i<cnt/k-1) printf("%05d\n",list[(i+2)*k-1].address);else{if(cnt%k==0) printf("-1\n");else{printf("%05d\n",list[(i+1)*k].address);for(int j=(i+1)*k;j<cnt-1;j++){printf("%05d %d %05d\n",list[j].address,list[j].data,list[j+1].address);}printf("%05d %d -1",list[cnt-1].address,list[cnt-1].data);}}}return 0;
}

A1075(25 排序)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct node{int rank,id,total=0;vector<int> score;int passnum=0;bool isshown=false;
};
bool cmp(node a,node b){if(a.total!=b.total) return a.total>b.total;else if(a.passnum!=b.passnum) return a.passnum>b.passnum;else return a.id<b.id;
}
int main(){int n,k,m,id,index,grade;scanf("%d%d%d",&n,&k,&m);vector<node> v(n+1);for(int i=1;i<=n;i++) v[i].score.resize(k+1,-1);vector<int> p(k+1);for(int i=1;i<=k;i++) scanf("%d",&p[i]);for(int i=0;i<m;i++){scanf("%d%d%d",&id,&index,&grade);v[id].id=id;v[id].score[index]=max(grade, v[id].score[index]);if(grade!=-1) v[id].isshown=true;else if(v[id].score[index]==-1) v[id].score[index]=-2;}for(int i=1;i<=n;i++){for(int j=1;j<=k;j++){if(v[i].score[j]!=-1&&v[i].score[j]!=-2) v[i].total+=v[i].score[j];if(v[i].score[j]==p[j]) v[i].passnum++;}}sort(v.begin()+1,v.end(),cmp);v[1].rank=1;for(int i=2;i<=n;i++){if(v[i].total==v[i-1].total) v[i].rank=v[i-1].rank;else v[i].rank=i;}for(int i=1;i<=n;i++){if(v[i].isshown==true){printf("%d %05d %d",v[i].rank,v[i].id,v[i].total);for(int j=1;j<=k;j++) {if(v[i].score[j]==-1) printf(" -");else if(v[i].score[j]==-2) printf(" 0");else printf(" %d",v[i].score[j]);}printf("\n");}}return 0;
}

A1076(30 图的bfs、记录给定最大层数上的结点和)

#include<cstdio>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
int n,l,k;
bool inq[1010];
vector<vector<int> > v;
struct node{int id,level;
};
int bfs(node a){fill(inq,inq+1010,false);int cnt=0;queue<node> q;q.push(a);inq[a.id]=true;while(!q.empty()){node t=q.front();q.pop();int tid=t.id;for(int i=0;i<v[tid].size();i++){int nextid=v[tid][i];if(inq[nextid]==false&&t.level<l){cnt++;node nextnode={nextid,t.level+1};q.push(nextnode);inq[nextid]=true;}}}return cnt;
}
int main(){scanf("%d%d",&n,&l);v.resize(n+1);int num,temp;for(int i=1;i<=n;i++){scanf("%d",&num);while(num--){scanf("%d",&temp);v[temp].push_back(i);}}scanf("%d",&k);while(k--){scanf("%d",&temp);node tnode={temp,0};printf("%d\n",bfs(tnode));}return 0;
}

A1077(20 多个字符串找公共后缀)

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main(){int n;scanf("%d\n",&n);string ans;int minlen;for(int i=0;i<n;i++){string s;getline(cin,s);reverse(s.begin(),s.end());if(i==0){ans=s;minlen=ans.length();}else{int len=s.length();minlen=min(minlen,len);for(int j=0;j<minlen;j++){if(s[j]!=ans[j]) {ans=s.substr(0,j);break;}}}}reverse(ans.begin(),ans.end());if(ans.size()==0) cout<<"nai";else cout<<ans;return 0;
}

A1078(25 hash 平方探查)

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
bool isprime(int n){if(n<=1) return false;int sqr=sqrt(1.0*n);for(int i=2;i<=sqr;i++){if(n%i==0) return false;}return true;
}
int main(){int msize,n,temp,cnt=0;scanf("%d%d",&msize,&n);while(!isprime(msize)) msize++;vector<int> v(msize,0);for(int i=0;i<n;i++){scanf("%d",&temp);int flag=0,ans;for(int j=0;j<msize;j++){int pos=(temp+j*j)%msize;if(v[pos]==0){v[pos]=temp;flag=1;ans=pos;break;}}if(cnt!=0) printf(" ");cnt++;if(flag) printf("%d",ans);else printf("-");}return 0;
}

A1079(25 树的dfs)

#include<cstdio>
#include<vector>
#include<cmath>
using namespace std;
int n;
double p,r,ans=0.0;
struct node{int data;vector<int> child;
}tree[100010];
void dfs(int index,int depth){if(tree[index].child.size()==0){ans+=tree[index].data*p*pow((1+r/100),depth);return;}for(int i=0;i<tree[index].child.size();i++) dfs(tree[index].child[i],depth+1);
}
int main(){scanf("%d %lf %lf",&n,&p,&r);int num,root,ischild[100010]={0},temp;for(int i=0;i<n;i++){scanf("%d",&num);if(num==0){scanf("%d",&temp);tree[i].data=temp;}else{for(int j=0;j<num;j++) {scanf("%d",&temp);ischild[temp]=1;tree[i].child.push_back(temp);} }}for(int i=0;i<n;i++){if(ischild[i]==0){root=i;break;}}dfs(root,0);printf("%.1f",ans);return 0;
}

A1080(30 排序)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct node{int ge,gi,sum,rank,id;vector<int> choice;
};
bool cmp1(node& a,node& b){if(a.sum!=b.sum) return a.sum>b.sum;else return a.ge>b.ge;
}
bool cmp2(node& a,node& b){return a.id<b.id;
}
int main(){int n,m,k;scanf("%d%d%d",&n,&m,&k);vector<int> sch_need(m);vector<node> stu(n);for(int i=0;i<m;i++) scanf("%d",&sch_need[i]);for(int i=0;i<n;i++){scanf("%d%d",&stu[i].ge,&stu[i].gi);stu[i].sum=stu[i].ge+stu[i].gi;stu[i].id=i;stu[i].choice.resize(k);for(int j=0;j<k;j++) scanf("%d",&stu[i].choice[j]);}sort(stu.begin(),stu.end(),cmp1);vector<node> ans[m];for(int i=0;i<n;i++){for(int j=0;j<k;j++){int temp=stu[i].choice[j];if(ans[temp].size()<sch_need[temp]||(ans[temp][ans[temp].size()-1].sum==stu[i].sum&&ans[temp][ans[temp].size()-1].ge==stu[i].ge)) {ans[temp].push_back(stu[i]);break;}}}for(int i=0;i<m;i++){sort(ans[i].begin(),ans[i].end(),cmp2);for(int j=0;j<ans[i].size();j++){if(j!=0) printf(" ");printf("%d",ans[i][j].id);}printf("\n");   }return 0;
}

A1081(20 分数相加)

方法一:
#include<cstdio>
long long gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);
}
int main() {int n;scanf("%d", &n);long long a2 = 0, b2 = 1,gcdvalue;for (int i = 0; i < n; i++) {long long a, b;scanf("%lld/%lld", &a, &b);gcdvalue = gcd(a, b);a /= gcdvalue;b /= gcdvalue;a2 = a * b2 + b * a2;b2 = b * b2;gcdvalue = gcd(a2, b2);a2 /= gcdvalue;b2 /= gcdvalue;}long long integer = a2 / b2;a2 = a2 - integer * b2;if (integer != 0) {printf("%lld", integer);if (a2 > 0) printf(" %lld/%lld",a2,b2);else if (a2 < 0) printf(" %lld/%lld", -a2, b2);}if (integer == 0 && a2 != 0) printf("%lld/%lld", a2, b2);if (integer == 0 && a2 == 0) printf("0");return 0;
}
方法二:
#include<cstdio>
#include<algorithm>
using namespace std;
struct node{long long up,down;
};
long long gcd(long long a,long long b){return b==0? abs(a) : gcd(b,a%b);
}
node reduction(node res){if(res.down<0){res.up=-res.up;res.down=-res.down;}if(res.up==0) res.down=1;else{long long d=gcd(abs(res.up),abs(res.down));res.up/=d;res.down/=d;}return res;
}
node add(node f1,node f2){node res;res.up=f1.down*f2.up+f1.up*f2.down;res.down=f1.down*f2.down;return reduction(res);
}
void showres(node res){res=reduction(res);if(res.down==1) printf("%lld\n",res.up);else if(abs(res.up)>abs(res.down)){printf("%lld %lld/%lld\n",res.up/res.down,abs(res.up%res.down),res.down);}else printf("%lld/%lld\n",res.up,res.down);
}
int main(){int n;scanf("%d",&n);node temp,ans;ans.up=0,ans.down=1;for(int i=0;i<n;i++){scanf("%lld/%lld",&temp.up,&temp.down);ans=add(ans,temp);} showres(ans);return 0;
}

A1082(25 分段处理字符串)

#include<iostream>
#include<string>
using namespace std;
int main(){string n;cin>>n;int len=n.length();int left=0,right=len-1;string num[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};string wei[5]={"Shi","Bai","Qian","Wan","Yi"};if(n[0]=='-') {cout<<"Fu";left++;}while(left+4<=right){right-=4;}while(left<len){bool flag=false;bool hasprint=false;while(left<=right){if(left>0&&n[left]=='0') flag=true;else{if(flag==true){cout<<" "<<"ling";flag=false;}if(left>0) cout<<" ";cout<<num[n[left]-'0'];hasprint=true;if(left!=right) cout<<" "<<wei[right-left-1];}left++;}if(right!=len-1&&hasprint==true)   cout<<" "<<wei[(len-1-right)/4+2]; right+=4; }return 0;
}

A1083(25 排序)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
struct node{string name,id;int grade;
}stu[1000];
bool cmp(node a,node b){return a.grade>b.grade;}
int main(){int n,score,g1,g2;string name,id;vector<node> ans;cin>>n;for(int i=0;i<n;i++) cin>>stu[i].name>>stu[i].id>>stu[i].grade;cin>>g1>>g2;for(int i=0;i<n;i++) {if(stu[i].grade>=g1&&stu[i].grade<=g2){ans.push_back(stu[i]);}}if(ans.size()==0) cout<<"NONE";else{sort(ans.begin(),ans.end(),cmp);for(int i=0;i<ans.size();i++) printf("%s %s\n",ans[i].name.c_str(),ans[i].id.c_str());} return 0;
}

A1084(20 字符串)

#include<iostream>
#include<string>
#include<cctype>
using namespace std;
int main(){string s1,s2,ans;cin>>s1>>s2;for(int i=0;i<s1.length();i++){if(s2.find(s1[i])==string::npos&&ans.find(toupper(s1[i]))==string::npos)ans+=toupper(s1[i]); }cout<<ans;return 0;
}

A1085(25 二分 twopoints 多解法)

二分法
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int main(){int n,p,ans=0;cin>>n>>p;vector<int> v(n);for(int i=0;i<n;i++)  cin>>v[i];sort(v.begin(),v.end());for(int i=0;i<n;i++){int j=upper_bound(v.begin()+i+1,v.begin()+n,(long long)p*v[i])-v.begin();ans=max(ans,j-i);}cout<<ans;return 0;
}two points
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn = 100010;
int main() {int n, p,num[maxn];scanf("%d%d", &n, &p);for (int i = 0; i < n; i++) {scanf("%d", &num[i]);}sort(num,num + n);int count=0,i=0,j=0;while(i<n&&j<n){while(j<n&&num[j]<=(long long)num[i]*p){count=max(count,j-i+1);j++;}i++;}printf("%d",count);return 0;
}

A1086(25 栈模拟中序和先序建树，输出后序)

#include<cstdio>
#include<stack>
#include<cstring>
using namespace std;
const int maxn=31;
int n,pre[maxn],in[maxn],num=0;
struct node{int data;node*left,*right;
};
node*create(int prel,int prer,int inl,int inr){if(prel>prer) return NULL;node*root=new node;root->data=pre[prel];int k=inl;while(in[k]!=root->data) k++;int leftnum=k-inl;root->left=create(prel+1,prel+leftnum,inl,k-1);root->right=create(prel+leftnum+1,prer,k+1,inr);return root;
}
void postorder(node*root){if(root==NULL) return;postorder(root->left);postorder(root->right);printf("%d",root->data);num++;if(num<n) printf(" ");
}
int main(){scanf("%d",&n);char s[5];stack<int> st;int temp,preindex=0,inindex=0;for(int i=0;i<2*n;i++){scanf("%s",s);if(strcmp(s,"Push")==0){scanf("%d",&temp);st.push(temp);pre[preindex++]=temp;}else{temp=st.top();st.pop();in[inindex++]=temp;}}node*root=create(0,n-1,0,n-1);postorder(root);return 0;
}

A1087(30 dijkst+dfs,边权点权、总边权、平均点权)

#include<iostream>
#include<map>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=210;
const int INF=0x7fffffff;
int n,k,G[maxn][maxn],d[maxn],happ[maxn],maxhapp=0,num=0;
double maxavg=0;
vector<int> pre[maxn];
bool vis[maxn]={false};
map<string,int> cityint;
map<int,string> citystr;
vector<int> temppath,path;
void dijkst(int s){fill(vis,vis+maxn,false);fill(d,d+maxn,INF);d[s]=0;for(int i=0;i<n;i++){int u=-1,MIN=INF;for(int j=0;j<n;j++){if(vis[j]==false&&d[j]<MIN){u=j;MIN=d[j];}}if(u==-1) return;vis[u]=true;for(int v=0;v<n;v++){if(vis[v]==false&&G[u][v]!=INF){if(d[u]+G[u][v]<d[v]){d[v]=G[u][v]+d[u];pre[v].clear();pre[v].push_back(u);}else if(d[u]+G[u][v]==d[v]) pre[v].push_back(u);}}}
}
void dfs(int s){if(s==0){num++;int temphapp=0;temppath.push_back(s);for(int i=temppath.size()-2;i>=0;i--) temphapp+=happ[temppath[i]];double tempavg=temphapp*1.0/(temppath.size()-1);if(temphapp>maxhapp){path=temppath;maxhapp=temphapp;maxavg=tempavg;}else if(temphapp==maxhapp&&tempavg>maxavg){path=temppath;maxavg=tempavg;}temppath.pop_back();return;}temppath.push_back(s);for(int i=0;i<pre[s].size();i++) dfs(pre[s][i]);temppath.pop_back();
}
int main(){string c1,c2;int a,index=0;cin>>n>>k>>c1;cityint[c1]=index++;citystr[cityint[c1]]=c1;for(int i=0;i<n-1;i++){cin>>c1>>a;cityint[c1]=index++;citystr[cityint[c1]]=c1;happ[cityint[c1]]=a;}fill(G[0],G[0]+maxn*maxn,INF);while(k--){cin>>c1>>c2>>a;G[cityint[c1]][cityint[c2]]=a;G[cityint[c2]][cityint[c1]]=a;}dijkst(0);dfs(cityint["ROM"]);printf("%d %d %d %d\n",num,d[cityint["ROM"]],maxhapp,(int)maxavg);for(int i=path.size()-1;i>=0;i--){cout<<citystr[path[i]];if(i>0) printf("->");}return 0;
}

A1088(20 分数的加减乘除)

#include<cstdio>
#include<algorithm>
using namespace std;
struct node{long long up, down;
};
long long gcd(long long a,long long b){return b==0? abs(a) : gcd(b,a%b);
}
node reduction(node res){if(res.down<0){res.up=-res.up;res.down=-res.down;}if(res.up==0) res.down=1;else{long long d=gcd(abs(res.up),abs(res.down));res.up/=d;res.down/=d;}return res;
}
node add(node f1,node f2){//分数相加node res;res.up=f1.down*f2.up+f1.up*f2.down;res.down=f1.down*f2.down;return reduction(res);
}
node minu(node f1,node f2){//分数相减node res;res.up=f1.up*f2.down-f2.up*f1.down;res.down=f1.down*f2.down;return reduction(res);
}
node multi(node f1,node f2){//分数相乘node res;res.up=f1.up*f2.up;res.down=f1.down*f2.down;return reduction(res);
}
node divide(node f1,node f2){//分数相除node res;res.up=f1.up*f2.down;res.down=f1.down*f2.up;return reduction(res);
}
void showres(node res){res=reduction(res);if(res.up<0) printf("(");if(res.down==1) printf("%lld",res.up);else if(abs(res.up)>abs(res.down)){printf("%lld %lld/%lld",res.up/res.down,abs(res.up%res.down),res.down);}else printf("%lld/%lld",res.up,res.down);if(res.up<0) printf(")");
}
int main(){node a,b,ans;scanf("%lld/%lld %lld/%lld",&a.up,&a.down,&b.up,&b.down);showres(a);printf(" + ");showres(b);printf(" = ");showres(add(a,b));printf("\n");showres(a);printf(" - ");showres(b);printf(" = ");showres(minu(a,b));printf("\n");showres(a);printf(" * ");showres(b);printf(" = ");showres(multi(a,b));printf("\n");showres(a);printf(" / ");showres(b);printf(" = ");if(b.up==0) printf("Inf");else showres(divide(a,b));printf("\n");return 0;
}

A1089 (25 插入排序和归并排序)

#include<iostream>
#include<algorithm>
using namespace std;
int main(){int a[100],b[100],n,j,m;cin>>n;for(int i=0;i<n;i++) cin>>a[i];for(int i=0;i<n;i++) cin>>b[i];for(j=0;j<n-1&&b[j]<=b[j+1];j++);for(m=j+1;m<n&&a[m]==b[m];m++);if(m==n){cout<<"Insertion Sort\n";sort(a,a+j+2);}else{cout<<"Merge Sort\n";int flag=1,k=1;while(flag){flag=0;for(int i=0;i<n;i++){if(a[i]!=b[i]) flag=1;}k=k*2;for(int i=0;i<n/k;i++) sort(a+i*k,a+(i+1)*k);sort(a+(n/k)*k,a+n);}}for(int i=0;i<n;i++){if(i!=0) cout<<" ";cout<<a[i];}return 0;
}

A1090(25 树的dfs求深度)

#include<cstdio>
#include<vector>
#include<cmath>
using namespace std;
int maxdepth=-1,num,n;
double p,r,ans=0;
vector<int> tree[100010];
void dfs(int index,int depth){if(tree[index].size()==0){if(depth>maxdepth){maxdepth=depth;num=1;}else if(depth==maxdepth) num++;return;}for(int i=0;i<tree[index].size();i++) dfs(tree[index][i],depth+1);
}
int main(){scanf("%d %lf %lf",&n,&p,&r);int father,root;for(int i=0;i<n;i++){scanf("%d",&father);if(father==-1) root=i;else tree[father].push_back(i);}dfs(root,0);ans=p*pow((1+r/100),maxdepth);printf("%.2f %d",ans,num);return 0;
}

A1091(30 经典BFS求矩阵中块的个数)

#include<cstdio>
#include<queue>
using namespace std;
struct node{int x,y,z;
}Node;
int m,n,L,t,ans=0;
bool inq[70][1300][140]={false};
int pix[70][1300][140]={0};
int Z[6]={0,0,0,0,1,-1};
int X[6]={0,0,1,-1,0,0};
int Y[6]={1,-1,0,0,0,0};
bool judge(int z,int y,int x){if(x>=n||x<0||y>=m||y<0||z<0||z>=L) return false;if(pix[z][y][x]==0||inq[z][y][x]==true) return false;return true;
}
int bfs(int z,int y,int x){int total=0;queue<node> q;Node.x=x,Node.y=y,Node.z=z;q.push(Node);inq[z][y][x]=true;while(!q.empty()){node Top=q.front();q.pop();total++;//注意在出队的时候计数 for(int i=0;i<6;i++){int newz=Top.z+Z[i];int newx=Top.x+X[i];int newy=Top.y+Y[i];if(judge(newz,newy,newx)){Node.x=newx,Node.y=newy,Node.z=newz;q.push(Node);inq[newz][newy][newx]=true;}}}if(total>=t) return total;else return 0;
}
int main(){scanf("%d%d%d%d",&m,&n,&L,&t);for(int i=0;i<L;i++){for(int j=0;j<m;j++){for(int k=0;k<n;k++){scanf("%d",&pix[i][j][k]);}}}for(int i=0;i<L;i++){for(int j=0;j<m;j++){for(int k=0;k<n;k++){if(inq[i][j][k]==false&&pix[i][j][k]==1) ans+=bfs(i,j,k);}}}printf("%d",ans);return 0;
}

A1092(字符串 hash)

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){string s1,s2;cin>>s1>>s2;vector<int> v(128,0);bool can=true;int miss=0;for(int i=0;i<s1.length();i++)  v[s1[i]]++;for(int i=0;i<s2.length();i++){v[s2[i]]-=1;if(v[s2[i]]<0) {can=false;++miss;}}if(can==false) cout<<"No "<<miss;else cout<<"Yes "<<s1.length()-s2.length();return 0;
}

A1093(25 逻辑题)

#include<iostream>
#include<string>
using namespace std;
int main(){string s;cin>>s;long long ans=0;int num1[100010]={0},num2[100010]={0};for(int i=1;i<s.length();i++){if(s[i-1]=='P') num1[i]=num1[i-1]+1;else num1[i]=num1[i-1];}for(int i=s.length()-2;i>=0;i--){if(s[i+1]=='T') num2[i]=num2[i+1]+1;else num2[i]=num2[i+1];}for(int i=1;i<s.length()-1;i++){if(s[i]=='A') ans+=num1[i]*num2[i];}cout<<ans%1000000007;return 0;
}

A1094(25 树的dfs)

#include<cstdio>
#include<vector>
using namespace std;
int n,m,level[100]={0},maxdepth=-1,ansnum=-1,anslevel;
vector<int> tree[100];
void dfs(int index,int depth){level[depth]++;if(tree[index].size()==0){if(maxdepth<depth) maxdepth=depth;return;}for(int i=0;i<tree[index].size();i++) dfs(tree[index][i],depth+1);
}
int main(){int id,k,temp;scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){scanf("%d%d",&id,&k);for(int j=0;j<k;j++){scanf("%d",&temp);tree[id].push_back(temp);}}dfs(1,1);for(int i=1;i<=maxdepth;i++){if(level[i]>ansnum){ansnum=level[i];anslevel=i;}}printf("%d %d",ansnum,anslevel);return 0;
}

A1095(30 校园停车 排序 两辆配对 时间查询输出)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
using namespace std;
struct node{string id;int time,flag=0;
};
bool cmp1(node &a,node &b){return a.id!=b.id? a.id<b.id : a.time<b.time;
}
bool cmp2(node &a,node &b){return a.time<b.time;
}
int main(){int n,m,hh,mm,ss,maxparking=-1,tempindex=0;string status;scanf("%d%d",&n,&m);vector<node> record(n),car;for(int i=0;i<n;i++){cin>>record[i].id;scanf("%d:%d:%d",&hh,&mm,&ss);cin>>status;record[i].time=hh*3600+mm*60+ss;record[i].flag=(status=="in")? 1 : -1;}sort(record.begin(),record.end(),cmp1);map<string,int> mp;for(int i=1;i<n;i++){if(record[i-1].flag==1&&record[i].flag==-1&&record[i-1].id==record[i].id){car.push_back(record[i-1]);car.push_back(record[i]);mp[record[i].id]+=record[i].time-record[i-1].time;if(maxparking<mp[record[i].id]) maxparking=mp[record[i].id];}}sort(car.begin(),car.end(),cmp2);vector<int> cnt(n);for(int i=0;i<car.size();i++){if(i==0) cnt[i]+=car[i].flag;else cnt[i]=cnt[i-1]+car[i].flag;}for(int i=0;i<m;i++){scanf("%d:%d:%d",&hh,&mm,&ss);int query=hh*3600+mm*60+ss;int j;for(j=tempindex;j<car.size();j++){if(query<car[j].time){printf("%d\n",cnt[j-1]);break;}else if(j==car.size()-1) printf("%d\n",cnt[j]);}tempindex=j;}for(auto it=mp.begin();it!=mp.end();it++){if(it->second==maxparking) cout<<it->first<<" ";}printf("%02d:%02d:%02d",maxparking/3600,maxparking%3600/60,maxparking%60);return 0;
}

A1096(20 找出连续的因数)

#include<iostream>
#include<cmath>
long long n,temp=1;
int main(){scanf("%d",&n);int st=0,len=0;for(int i=2;i<=sqrt(n*1.0);i++){int j,temp=1;for(j=i;j<=sqrt(n*1.0);j++){temp*=j;if(n%temp!=0) break;}if(j-i>len){len=j-i;st=i;}}if(st==0) printf("1\n%d",n);else{printf("%d\n",len);for(int i=0;i<len;i++){if(i!=0) printf("*");printf("%d",st+i);}}return 0;
}

A1097(25 链表分离)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{int address,data,next;
}list[100010];
int main(){int begin,n,address;scanf("%d%d",&begin,&n);vector<int> flag(10010,0);vector<node> remain,remov;for(int i=0;i<n;i++){scanf("%d",&address);scanf("%d%d",&list[address].data,&list[address].next);list[address].address=address;}while(begin!=-1){if(flag[abs(list[begin].data)]==0) {flag[abs(list[begin].data)]=1;remain.push_back(list[begin]);}else remov.push_back(list[begin]);begin=list[begin].next;}for(int i=0;i<remain.size()-1;i++) printf("%05d %d %05d\n",remain[i].address,remain[i].data,remain[i+1].address);      printf("%05d %d -1\n",remain[remain.size()-1].address,remain[remain.size()-1].data);if(remov.size()!=0){for(int i=0;i<remov.size()-1;i++) printf("%05d %d %05d\n",remov[i].address,remov[i].data,remov[i+1].address);       printf("%05d %d -1",remov[remov.size()-1].address,remov[remov.size()-1].data);}return 0;
}

A1098(25 判断插入排序和堆排序)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
void downadjust(vector<int>&b,int low,int high){int i=low,j=i*2;while(j<=high){if(j+1<=high&&b[j+1]>b[j]) j=j+1;if(b[i]>=b[j]) break;swap(b[i],b[j]);i=j;j=i*2;}
}
int main(){int n,p,q;scanf("%d",&n);vector<int>a(n+1),b(n+1);for(int i=1;i<=n;i++) scanf("%d",&a[i]);for(int i=1;i<=n;i++) scanf("%d",&b[i]);for(p=2;b[p-1]<=b[p]&&p<=n;p++);int index=p;while(index<=n&&a[index]==b[index]) index++;if(index==n+1){printf("Insertion Sort\n");sort(b.begin()+1,b.begin()+p+1); }else{printf("Heap Sort\n");for(q=n;b[q]>=b[1]&&q>2;q--);swap(b[1],b[q]);downadjust(b,1,q-1);}for(int i=1;i<=n;i++){if(i>1) printf(" ");printf("%d",b[i]);}return 0;
}

A1099(30 中序遍历建立二叉搜索树)

#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=101;
struct node{int v;int left,right;
}Node[maxn];
int n,a[maxn],index=0,num=0;
void inorder(int root){if(root==-1) return;inorder(Node[root].left);Node[root].v=a[index++];inorder(Node[root].right);
}
void bfs(int root){queue<int> q;q.push(root);while(!q.empty()){int temp=q.front();q.pop();if(num>0) printf(" ");printf("%d",Node[temp].v);num++;if(Node[temp].left!=-1) q.push(Node[temp].left);if(Node[temp].right!=-1) q.push(Node[temp].right);}
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d%d",&Node[i].left,&Node[i].right);for(int i=0;i<n;i++) scanf("%d",&a[i]);sort(a,a+n);inorder(0);bfs(0);return 0;
}

A1100(20 字符串 string map)

#include<iostream>
#include<string>
#include<map>
using namespace std;
string unitdigit[13]={"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug",
"sep", "oct", "nov", "dec"};
string tendigit[13]= {"tret", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo",
"syy", "lok", "mer", "jou"};
string numtostr[169];
map<string,int> strtonum;
void init(){for(int i=0;i<13;i++){numtostr[i]=unitdigit[i];strtonum[unitdigit[i]]=i;numtostr[i*13]=tendigit[i];strtonum[tendigit[i]]=i*13;}for(int i=1;i<13;i++){for(int j=1;j<13;j++){string str=tendigit[i]+" "+unitdigit[j];numtostr[i*13+j]=str;strtonum[str]=i*13+j;}}
}
int main(){init();int n,temp;scanf("%d\n",&n);string s;while(n--){getline(cin,s);if(s[0]>='0'&&s[0]<='9'){temp=stoi(s);cout<<numtostr[temp]<<endl;}else cout<<strtonum[s]<<endl;}return 0;
}

A1101(25 逻辑题)

#include<iostream>
#include<set>
using namespace std;
int main(){int n,a[100010],leftmax=-1,rightmin=0x7fffffff;int left[100010],right[100010];set<int> ans;cin>>n;for(int i=0;i<n;i++) cin>>a[i];left[0]=a[0];for(int i=1;i<n;i++){if(a[i-1]>leftmax) leftmax=a[i-1];left[i]=leftmax;}right[n-1]=a[n-1];for(int i=n-2;i>=0;i--){if(a[i+1]<rightmin) rightmin=a[i+1];right[i]=rightmin;}for(int i=0;i<n;i++){if(a[i]>=left[i]&&a[i]<=right[i]) ans.insert(a[i]);}cout<<ans.size()<<endl;for(auto it=ans.begin();it!=ans.end();it++){if(it!=ans.begin()) cout<<" ";cout<<*it;}cout<<endl;return 0;
}

A1102(25 二叉树反转)

#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct node{int left,right;
}tree[11];
int n,ischild[11]={0},root,num1=0,num2=0;
void layerorder(int root){queue<int> q;q.push(root);while(!q.empty()){int Top=q.front();q.pop();printf("%d",Top);num1++;if(num1<n) printf(" ");if(tree[Top].left!=-1) q.push(tree[Top].left);if(tree[Top].right!=-1) q.push(tree[Top].right); }
}
void inorder(int root){if(root==-1) return;inorder(tree[root].left);printf("%d",root);num2++;if(num2<n) printf(" ");inorder(tree[root].right);
}
void invert(int root){if(root==-1) return;invert(tree[root].left);invert(tree[root].right);swap(tree[root].left,tree[root].right);
}
int main(){scanf("%d",&n);char c1,c2;for(int i=0;i<n;i++){scanf("%*c%c %c",&c1,&c2);if(c1=='-') tree[i].left=-1;else {ischild[c1-'0']=1;tree[i].left=c1-'0';}if(c2=='-') tree[i].right=-1;else {ischild[c2-'0']=1;tree[i].right=c2-'0';}}for(int i=0;i<n;i++){if(ischild[i]==0){root=i;break;} }invert(root);layerorder(root);printf("\n");inorder(root);return 0;
}

A1103(30 DFS 分解因式)

#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
int n,k,p,maxfac=-1;
vector<int> fac,temp,ans;
void init(){int i=0;while(pow(i,p)<=n){fac.push_back(pow(i,p));i++;}
}
void dfs(int index,int nowk,int sum,int facsum){if(nowk==k&&sum==n){if(facsum>maxfac){maxfac=facsum;ans=temp;}return;}if(nowk>k||sum>n) return;if(index>=1){temp.push_back(index);dfs(index,nowk+1,sum+fac[index],facsum+index);temp.pop_back();dfs(index-1,nowk,sum,facsum);}
}
int main(){scanf("%d%d%d",&n,&k,&p);init();dfs(fac.size()-1,0,0,0);if(maxfac==-1) printf("Impossible");else{printf("%d = ",n);for(int i=0;i<ans.size();i++){if(i>0) printf(" + ");printf("%d^%d",ans[i],p);}}return 0;
}

A1104(20 片段和 逻辑题)

#include<iostream>
using namespace std;
int main(){int n;long double ans=0,temp;cin>>n;for(int i=0;i<n;i++){cin>>temp;ans+=temp*(n-i)*(i+1);}printf("%.2llf\n",ans);return 0;
}

A1105(25 顺时针螺旋二维数组，注意判断下标越界)

#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
bool cmp(int a,int b) {return a>b;
}
int main() {int N, m,n;scanf("%d",&N);vector<int>t(N);for(int i=0; i<N; i++) scanf("%d",&t[i]);sort(t.begin(),t.end(),cmp);m=(int)(ceil(sqrt(N*1.0)));while(N%m!=0) m++;n=N/m;vector<vector<int> >v(m,vector<int>(n));int level=m/2+m%2,index=0;for(int i=0; i<level; i++) {for(int j=i; j<n-i&&index<N; j++)  v[i][j]=t[index++];for(int j=i+1; j<m-i&&index<N; j++) v[j][n-i-1]=t[index++];for(int j=n-i-2; j>=i&&index<N; j--) v[m-i-1][j]=t[index++];for(int j=m-i-2; j>i&&index<N; j--) v[j][i]=t[index++];}for(int i=0; i<m; i++) {if(i!=0) printf("\n");for(int j=0; j<n; j++) {if(j!=0) printf(" ");printf("%d",v[i][j]);}}return 0;
}

A1106(25 树的dfs)

#include<cstdio>
#include<vector>
#include<cmath>
using namespace std;
int n,num=0,mindepth=0x7fffffff;
double p,r,ans=0.0;
vector<int> tree[100010];
void dfs(int index,int depth){if(tree[index].size()==0){if(mindepth>depth){mindepth=depth;num=1;}else if(mindepth==depth) num++;return;}for(int i=0;i<tree[index].size();i++) dfs(tree[index][i],depth+1);
}
int main(){scanf("%d %lf %lf",&n,&p,&r);int k,temp;for(int i=0;i<n;i++){scanf("%d",&k);for(int j=0;j<k;j++){scanf("%d",&temp);tree[i].push_back(temp);}}dfs(0,0);ans=p*pow((1+r/100),mindepth);printf("%.4f %d",ans,num);return 0;
}

A1107(30 经典并查集)

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>father,isroot;
bool cmp(int a,int b){return a>b;}
int findfather(int x){int a=x;while(x!=father[x]) x=father[x];while(a!=father[a]){int z=a;a=father[a];father[z]=x;}return x;
}
void Union(int a,int b){int faA=findfather(a);int faB=findfather(b);if(faA!=faB) father[faA]=faB;
}
int main(){int course[1001]={0},n,num,courseid,cnt=0;scanf("%d",&n);father.resize(n+1);isroot.resize(n+1,0);for(int i=1;i<=n;i++) father[i]=i;for(int i=1;i<=n;i++){scanf("%d:",&num);while(num--){scanf("%d",&courseid);if(course[courseid]==0) course[courseid]=i;Union(i,findfather(course[courseid])); }}for(int i=1;i<=n;i++) isroot[findfather(i)]++;for(int i=1;i<=n;i++){if(isroot[i]!=0) cnt++;}sort(isroot.begin(),isroot.end(),cmp);printf("%d\n",cnt);for(int i=0;i<cnt;i++){if(i>0) printf(" ");printf("%d",isroot[i]);}return 0;
}

A1108(20 字符串 sscanf、sprintf)

#include<cstdio>
#include<cstring>
int main(){int n,num=0;double sum=0;scanf("%d",&n);for(int i=0;i<n;i++){char s[50],b[50];double temp;int flag=0;scanf("%s",s);sscanf(s,"%lf",&temp);sprintf(b,"%.2f",temp);for(int j=0;j<strlen(s);j++){if(s[j]!=b[j]) flag=1;}if(flag||temp<-1000||temp>1000){printf("ERROR: %s is not a legal number\n",s);continue;}num++;sum+=temp;}if(num==0) printf("The average of 0 numbers is Undefined");else if(num==1) printf("The average of 1 number is %.2f",sum);else printf("The average of %d numbers is %.2f",num,sum/num);return 0;
}

A1109(25 排序)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
struct node{int height;char name[10];
}v[10010];
bool cmp(node a,node b){return a.height!=b.height? a.height>b.height : strcmp(a.name,b.name)<0;
}
int main(){int n,k,index=0,num;scanf("%d%d",&n,&k);int m=round(n*1.0/k);for(int i=0;i<n;i++) scanf("%s%d",v[i].name,&v[i].height);sort(v,v+n,cmp);for(int i=0;i<k;i++){if(i==0) num=n-m*(k-1);else num=m;vector<node> temp(num);temp[num/2]=v[index++];for(int j=0;j<num;j++){if(num/2-j-1>=0) temp[num/2-j-1]=v[index++];if(num/2+j+1<num) temp[num/2+j+1]=v[index++];}for(int j=0;j<num;j++){if(j>0) printf(" ");printf("%s",temp[j].name);}printf("\n");}return 0;
}

A1110(25 判断是否是完全二叉树 dfs)

#include<iostream>
#include<string>
using namespace std;
struct node{int left,right;
}v[30];
int n,maxn=-1,ans;
void dfs(int root,int index){if(index>maxn){maxn=index;ans=root;}if(v[root].left!=-1) dfs(v[root].left,index*2);if(v[root].right!=-1) dfs(v[root].right,index*2+1);
}
int main(){scanf("%d",&n);string l,r;int ischild[30]={0},root=0;for(int i=0;i<n;i++){cin>>l>>r;if(l=="-") v[i].left=-1;else {v[i].left=stoi(l);ischild[v[i].left]=1;}if(r=="-") v[i].right=-1;else {v[i].right=stoi(r);ischild[v[i].right]=1;}}while(ischild[root]==1) root++;dfs(root,1);if(maxn==n) printf("YES %d",ans);else printf("NO %d",root);return 0;
}

A1111(30 两次dijkst加dfs,最短、最快路径、结点最少路径)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int INF=0x7fffffff;
const int maxn=520;
int n,m,st,ed,e[maxn][maxn],t[maxn][maxn],d[maxn],time[maxn],dispre[maxn],timepre[maxn],fast[maxn],fewnum[maxn];
bool vis[maxn];
vector<int> dispath,timepath;
void dis_dijkst(int s){fill(vis,vis+maxn,false);fill(d,d+maxn,INF);fill(fast,fast+maxn,INF);for(int i=0;i<n;i++) dispre[i]=i;d[s]=0;fast[s]=0;for(int i=0;i<n;i++){int u=-1,MIN=INF;for(int j=0;j<n;j++){if(vis[j]==false&&d[j]<MIN){u=j;MIN=d[j];}}if(u==-1) return;vis[u]=true;for(int v=0;v<n;v++){if(vis[v]==false&&e[u][v]!=INF){if(d[u]+e[u][v]<d[v]){d[v]=d[u]+e[u][v];fast[v]=fast[u]+t[u][v];dispre[v]=u;}else if(d[u]+e[u][v]==d[v]&&fast[v]>fast[u]+t[u][v]){fast[v]=fast[u]+t[u][v];dispre[v]=u;}}} }
}
void time_dijkst(int s){fill(time,time+maxn,INF);fill(vis,vis+maxn,false);time[s]=0;fewnum[s]=1;for(int i=0;i<n;i++) timepre[i]=i;for(int i=0;i<n;i++){int u=-1,MIN=INF;for(int j=0;j<n;j++){if(vis[j]==false&&time[j]<MIN){u=j;MIN=time[j];}}if(u==-1) return;vis[u]=true;for(int v=0;v<n;v++){if(vis[v]==false&&t[u][v]!=INF){if(time[v]>time[u]+t[u][v]){time[v]=time[u]+t[u][v];timepre[v]=u;fewnum[v]=fewnum[u]+1;}else if(time[v]==time[u]+t[u][v]&&fewnum[v]>fewnum[u]+1){timepre[v]=u;fewnum[v]=fewnum[u]+1;}}}}
}
void dis_dfs(int ed){dispath.push_back(ed);if(ed==st) return;dis_dfs(dispre[ed]);
}
void time_dfs(int ed){timepath.push_back(ed);if(ed==st) return;time_dfs(timepre[ed]);
}
int main(){scanf("%d%d",&n,&m);int v1,v2,flag,len,ti;fill(e[0],e[0]+maxn*maxn,INF);fill(t[0],t[0]+maxn*maxn,INF);for(int i=0;i<m;i++){scanf("%d%d%d%d%d",&v1,&v2,&flag,&len,&ti);e[v1][v2]=len;t[v1][v2]=ti;if(flag!=1){e[v2][v1]=len;t[v2][v1]=ti;}}scanf("%d%d",&st,&ed);dis_dijkst(st);time_dijkst(st);dis_dfs(ed);time_dfs(ed);if(dispath==timepath){printf("Distance = %d; Time = %d: ",d[ed],time[ed]);for(int i=dispath.size()-1;i>=0;i--){printf("%d",dispath[i]);if(i>0) printf(" -> ");}}else{printf("Distance = %d: ",d[ed]);for(int i=dispath.size()-1;i>=0;i--){printf("%d",dispath[i]);if(i>0) printf(" -> ");}printf("\n");printf("Time = %d: ",time[ed]);for(int i=timepath.size()-1;i>=0;i--){printf("%d",timepath[i]);if(i>0) printf(" -> ");}}return 0;
}

A1112(20 字符串、找出“坏键”)

#include<iostream>
#include<string>
#include<map>
using namespace std;
int main(){map<char,bool> mp,hasprint;string s,ans;int k,cnt=1;scanf("%d\n",&k);cin>>s;s="*"+s+"*";for(int i=1;i<s.length();i++){if(s[i]==s[i-1]) cnt++;else{if(cnt%k!=0) mp[s[i-1]]=true;cnt=1;}}for(int i=1;i<s.length()-1;i++){ans+=s[i];if(mp[s[i]]!=true){if(hasprint.count(s[i])!=true){cout<<s[i];hasprint[s[i]]=true;}i+=k-1;}}cout<<endl<<ans;return 0;
}

A1113(25 水题)

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int main(){int n,s=0,s1=0;scanf("%d",&n);vector<int>v(n);for(int i=0;i<n;i++) scanf("%d",&v[i]);sort(v.begin(),v.end());for(int i=0;i<n;i++){if(i<n/2) s1+=v[i];s+=v[i];}printf("%d %d",n%2==0? 0:1,s-s1*2);return 0;
}

A1114 (25 并查集)

#include<cstdio>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
int n,father[10001],family[10001],exist[10001];
struct node{int sets,area;
}v[10001];
struct fam{int id,num=0;double avgset=0, avgarea=0;
}f[10001];
bool cmp(fam a,fam b){return a.avgarea!=b.avgarea ? a.avgarea>b.avgarea : a.id<b.id;
}
int findfather(int x){int a=x;while(x!=father[x]) x=father[x];while(a!=father[a]){int z=a;a=father[a];father[z]=x;}return x;
}
void Union(int a,int b){int faA=findfather(a);int faB=findfather(b);if(faA>faB) father[faA]=faB;else if(faA<faB) father[faB]=faA;
}
int main(){set<int> ans;scanf("%d",&n);for(int i=0;i<10001;i++) father[i]=i;for(int i=0;i<n;i++){int id,fa,mo,k,childid;scanf("%d%d%d%d",&id,&fa,&mo,&k);exist[id]=1;if(fa!=-1){exist[fa]=1;Union(id,fa);} if(mo!=-1) {Union(id,mo);;exist[mo]=1;}while(k--){scanf("%d",&childid);exist[childid]=1;Union(id,childid);} scanf("%d%d",&v[id].sets,&v[id].area);}for(int i=0;i<10001;i++){if(exist[i]==1){ans.insert(findfather(i));f[findfather(i)].id=findfather(i);f[findfather(i)].num++;f[findfather(i)].avgset+=v[i].sets;f[findfather(i)].avgarea+=v[i].area;}}for(auto it=ans.begin();it!=ans.end();it++){f[*it].avgarea=f[*it].avgarea/f[*it].num;f[*it].avgset=f[*it].avgset/f[*it].num;} sort(f,f+10001,cmp);printf("%d\n",ans.size());for(int i=0;i<ans.size();i++) printf("%04d %d %.3f %.3f\n",f[i].id,f[i].num,f[i].avgset,f[i].avgarea);return 0;
}

A1115(30 二叉搜索树建立与DFS)

#include<cstdio>
struct node{int v;node*left,*right;
};
int maxlevel=-1,level=0,n1=0,n2=0;
void insert(node *&root,int v){level++;if(root==NULL){root=new node;root->v=v;root->left=root->right=NULL;if(maxlevel<level) maxlevel=level;level=0;return;}if(v<=root->v) insert(root->left,v);else insert(root->right,v);
}
void dfs(node*root,int level){if(root==NULL) return;if(level==maxlevel) n1++;else if(level==maxlevel-1) n2++;dfs(root->left,level+1);dfs(root->right,level+1);
}
int main(){int n,a;node*root=NULL;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&a);insert(root,a);}dfs(root,1);printf("%d + %d = %d",n1,n2,n1+n2);return 0;
}

A1116(20 水题)

#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
bool isprime(int x){if(x<=1) return false;int sqr=sqrt(1.0*x);for(int i=2;i<=sqr;i++){if(x%i==0) return false;}return true;
}
int main(){int n,temp,k;scanf("%d",&n);map<int,int> mp,hascheck;for(int i=1;i<=n;i++){scanf("%d",&temp);mp[temp]=i;}scanf("%d",&k);while(k--){scanf("%d",&temp);if(mp.count(temp)==0) printf("%04d: Are you kidding?\n",temp);else{if(hascheck.count(temp)==0){hascheck[temp]=1;if(mp[temp]==1) printf("%04d: Mystery Award\n",temp);else if(isprime(mp[temp])) printf("%04d: Minion\n",temp);else printf("%04d: Chocolate\n",temp);}else printf("%04d: Checked\n",temp);}}return 0;
}

A1117(25 逻辑题)

#include<iostream>
#include<algorithm>
using namespace std;
int main(){int a[100010]={0},n,j=0;scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&a[i]);sort(a,a+n,greater<int>());while(j<n&&a[j]>j+1) j++;printf("%d",j);return 0;
}

A1118(25 并查集)

#include<cstdio>
#include<set>
using namespace std;
int n,father[10006],tree[10006]={0};
int findfather(int x){int a=x;while(x!=father[x]) x=father[x];while(a!=father[a]){int z=a;a=father[a];father[z]=x;}return x;
}
void Union(int a,int b){int fa=findfather(a);int fb=findfather(b);if(fa!=fb) father[fa]=fb;
}
int main(){int k,q,t,temp,num=0;scanf("%d",&n);set<int> birds;for(int i=0;i<10006;i++) father[i]=i;for(int i=0;i<n;i++){scanf("%d%d",&k,&t);birds.insert(t);for(int j=0;j<k-1;j++){scanf("%d",&temp);birds.insert(temp);Union(temp,t);}}for(auto it=birds.begin();it!=birds.end();it++) tree[findfather(*it)]++;for(int i=0;i<10006;i++){if(tree[i]!=0) num++;}printf("%d %d\n",num,birds.size());scanf("%d",&q);while(q--){scanf("%d%d",&t,&temp);if(findfather(t)==findfather(temp)) printf("Yes\n");else printf("No\n");}return 0;
}

A1119 (30 树的前序和后序 求中序)

#include<cstdio>
#include<vector>
using namespace std;
int n,post[31],pre[30],num=0;
bool uniqu=true;
struct node{int v;node*left,*right;
};
node* create(int prel,int prer,int postl,int postr){if(prel>prer) return NULL;node*root=new node;root->v=pre[prel];//每次取先序序列的第一个为根节点int k=prel+1;while(k<=prer&&pre[k]!=post[postr-1]) k++;if(prel+1<n&&postr-1>=0&&pre[prel+1]==post[postr-1]) uniqu=false;//如果先序序列的第二个元素等于后序序列的倒数第二个元素，说明树的个数不唯一，因为无法确定该结点属于左子树还是右子树int numleft=k-prel-1;root->left=create(prel+1,k-1,postl,postl+numleft-1);root->right=create(k,prer,postl+numleft,postr-1);//如果树的个数不唯一，假设该不确定节点属于右子树return root;
}
void inorder(node*root){if(root==NULL) return;inorder(root->left);if(num>0) printf(" ");printf("%d",root->v);num++;inorder(root->right);
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&pre[i]);for(int i=0;i<n;i++) scanf("%d",&post[i]);node*root=NULL;root=create(0,n-1,0,n-1);if(uniqu) printf("Yes\n");else printf("No\n");inorder(root);printf("\n");return 0;
}

A1120(20 水题)

#include<iostream>
#include<set>
#include<string>
using namespace std;
const int maxn=10010;
int main(){int n,sum[maxn]={0};string str;set<int> ans;cin>>n;bool flag[maxn]={false};for(int i=0;i<n;i++){cin>>str;int len=str.size();for(int j=0;j<len;j++){sum[i]+=(str[j]-'0');}}for(int i=0;i<n;i++) ans.insert(sum[i]);printf("%d\n",ans.size());for(auto it=ans.begin();it!=ans.end();it++) {if(it!=ans.begin()) printf(" %d",*it);else printf("%d",*it);}return 0;
}

A1121 (25 map、set应用)

#include<cstdio>
#include<set>
#include<vector>
#include<map>
using namespace std;
bool flag[100000] = { false };
map<int, int>couple;
int main() {int n, m, p1, p2;set<int> ans;vector<int> guests;scanf("%d", &n);while (n--) {scanf("%d%d", &p1, &p2);couple[p1] = p2;couple[p2] = p1;}scanf("%d", &m);for (int i = 0; i < m; i++) {scanf("%d", &p1);guests.push_back(p1);flag[p1] = true;}for (int i = 0; i < guests.size(); i++) {if (couple.count(guests[i])==0||(flag[couple[guests[i]]]==false)&& couple.count(guests[i]) != 0) ans.insert(guests[i]);}printf("%d\n", ans.size());for (auto it = ans.begin(); it != ans.end(); it++) {if (it != ans.begin()) printf(" ");printf("%05d", *it);}return 0;
}

A1122(25 哈密顿回路 set)

#include<cstdio>
#include<set>
#include<vector>
using namespace std;
int main(){int n,m,k,c1,c2,g[210][210];scanf("%d%d",&n,&m);while(m--){scanf("%d%d",&c1,&c2);g[c1][c2]=1;g[c2][c1]=1;}scanf("%d",&k);while(k--){int num,flag1=1,flag2=1;scanf("%d",&num);vector<int> v(num);set<int> s;for(int i=0;i<num;i++){scanf("%d",&v[i]); s.insert(v[i]);}    if(num-1!=n||s.size()!=n||v[0]!=v[num-1]) flag1=0;for(int i=0;i<num-1;i++){if(g[v[i]][v[i+1]]!=1) flag2=0;}if(flag1==1&&flag2==1) printf("YES\n");else printf("NO\n");}return 0;
}

A1123(30 AVL 、层序遍历、判断是否是完全二叉树)

#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct node{int v,height;node* left,*right;
};
node* newnode(int v){node* Node=new node;Node->v=v;Node->left=Node->right=NULL;Node->height=1;return Node;
}
int getheight(node *root){if(root==NULL) return 0;return root->height;
}
void updateheight(node *root){root->height=max(getheight(root->left),getheight(root->right))+1;
}
int getbalance(node *root){return getheight(root->left)-getheight(root->right);
}
void R(node*&root){node*temp=root->left;root->left=temp->right;temp->right=root;updateheight(root);updateheight(temp);root=temp;
}
void L(node*&root){node*temp=root->right;root->right=temp->left;temp->left=root;updateheight(root);updateheight(temp);root=temp;
}
void insert(node*&root,int v){if(root==NULL){root=newnode(v);return;}if(root->v>v){insert(root->left,v);updateheight(root);if(getbalance(root)==2){if(getbalance(root->left)==1){R(root);}else if(getbalance(root->left)==-1){L(root->left);R(root);}}}else{insert(root->right,v);updateheight(root);if(getbalance(root)==-2){if(getbalance(root->right)==-1){L(root);}else if(getbalance(root->right)==1){R(root->right);L(root);}}}
}
int iscomplete=1,after=0,n,num=0;
void layerorder(node*root){queue<node*>q;q.push(root);while(!q.empty()){node* t=q.front();q.pop();if(num>0) printf(" ");printf("%d",t->v);num++;if(t->left!=NULL){if(after) iscomplete=0;q.push(t->left);}else after=1;if(t->right!=NULL){if(after) iscomplete=0;q.push(t->right);}else after=1;}
}
int main(){int temp;scanf("%d",&n);node*root=NULL;for(int i=0;i<n;i++){scanf("%d",&temp);insert(root,temp);}layerorder(root);if(iscomplete) printf("\nYES");else printf("\nNO");return 0;
}

A1124 20 逻辑题

#include<iostream>
#include<vector>
#include<string>
#include<map>
using namespace std;
int main(){int m,n,s;cin>>m>>n>>s;vector<string> v(m+1);map<string,bool> win;for(int i=1;i<=m;i++)    cin>>v[i];if(s>m) cout<<"Keep going...";else{for(int i=s;i<=m;i++){if(win.count(v[i])==0){win[v[i]]=true;cout<<v[i]<<endl;i+=n-1;}}}return 0;
}

A1125(25 贪心 排序)

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int main(){int n;scanf("%d",&n);vector<int> v(n);for(int i=0;i<n;i++) scanf("%d",&v[i]);sort(v.begin(),v.end());double seg=v[0];for(int i=0;i<n-1;i++) seg=(seg+v[i+1])/2;printf("%d",(int)seg);return 0;
}

A1126(25 欧拉图 欧拉回路 欧拉路径)

#include<cstdio>
int g[510][510],degree[510]={0},n,m;
bool vis[510]={false};
void dfs(int u){vis[u]=true;for(int i=1;i<=n;i++){if(vis[i]==false&&g[u][i]==1){dfs(i);}}
}
int main(){scanf("%d%d",&n,&m);int v1,v2,evennum=0,block=0;while(m--){scanf("%d%d",&v1,&v2);g[v1][v2]=g[v2][v1]=1;degree[v1]++;degree[v2]++;}for(int i=1;i<=n;i++){if(i>1) printf(" ");printf("%d",degree[i]);if(degree[i]%2==0) evennum++;}for(int i=1;i<=n;i++){if(vis[i]==false){dfs(i);block++;}   }if(block==1&&evennum==n) printf("\nEulerian");else if(block==1&&evennum==n-2) printf("\nSemi-Eulerian");else printf("\nNon-Eulerian");return 0;
}

A1127(30 中序后序建树，dfs，输出z字形层序遍历)

方法一:层序遍历用结构体统计每层结点
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
int n,post[31],in[31];
vector<int> level[35];
struct node{int v,depth;node* left,*right;
};
node* create(int postl,int postr,int inl,int inr){if(postl>postr) return NULL;node *root=new node;root->v=post[postr];int k=inl;while(in[k]!=root->v&&k<=inr) k++;int leftnum=k-inl;root->left=create(postl,postl+leftnum-1,inl,k-1);root->right=create(postl+leftnum,postr-1,k+1,inr);return root;
}
void bfs(node*root){queue<node> q;q.push(node{root->v,1,root->left,root->right});while(!q.empty()){node t=q.front();q.pop();level[t.depth].push_back(t.v);if(t.left!=NULL) q.push(node{t.left->v,t.depth+1,t.left->left,t.left->right});if(t.right!=NULL) q.push(node{t.right->v,t.depth+1,t.right->left,t.right->right});}
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&in[i]);for(int i=0;i<n;i++) scanf("%d",&post[i]);node*root=create(0,n-1,0,n-1);bfs(root);printf("%d",root->v);for(int i=2;i<35;i++){if(i%2==0){for(int j=0;j<level[i].size();j++) printf(" %d",level[i][j]);}else{for(int j=level[i].size()-1;j>=0;j--) printf(" %d",level[i][j]);}}return 0;
}方法二:层序遍历用当前队列的元素个数统计每层节点
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
int n,post[31],in[31],depth=1;
vector<int> level[31];
struct node{int v;node* left,*right;
};
node* create(int postl,int postr,int inl,int inr){if(postl>postr) return NULL;node *root=new node;root->v=post[postr];int k=inl;while(in[k]!=root->v&&k<=inr) k++;int leftnum=k-inl;root->left=create(postl,postl+leftnum-1,inl,k-1);root->right=create(postl+leftnum,postr-1,k+1,inr);return root;
}
void bfs(node*root){queue<node*> q;q.push(root);while(!q.empty()){int count=q.size();while(count--){node* t=q.front();q.pop();level[depth].push_back(t->v);if(t->left!=NULL) q.push(t->left);if(t->right!=NULL) q.push(t->right);}depth++;}
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&in[i]);for(int i=0;i<n;i++) scanf("%d",&post[i]);node*root=create(0,n-1,0,n-1);bfs(root);printf("%d",root->v);for(int i=2;i<=depth;i++){if(i%2==0){for(int j=0;j<level[i].size();j++) printf(" %d",level[i][j]);}else{for(int j=level[i].size()-1;j>=0;j--) printf(" %d",level[i][j]);}}return 0;
}

A1128(20 n皇后问题)

#include<cstdio>
#include<algorithm>
using namespace std;
int main(){int k,n,a[1005];scanf("%d",&k);while(k--){int flag=1;scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]);for(int i=2;i<=n;i++){for(int j=1;j<i;j++){if(a[j]==a[i]||abs(i-j)==abs(a[j]-a[i])){flag=0;break;}}if(flag==0){printf("NO\n");break;}}if(flag) printf("YES\n");}return 0;
}

A1129 (25 set的应用 结构体内运算符重载)

#include<cstdio>
#include<vector>
#include<set>
#include<algorithm>
using namespace std;
struct node{int id,fre;
};
bool operator < (const node &a,const node &b){return a.fre!=b.fre? a.fre>b.fre : a.id<b.id;
}
int main(){int n,k,item,times[50005]={0};scanf("%d%d",&n,&k);set<node> s; for(int i=0;i<n;i++){scanf("%d",&item);if(i!=0){int j=0;printf("%d:",item);for(auto it=s.begin();j<k&&it!=s.end();it++,j++) printf(" %d",*it);printf("\n");}auto it=s.find(node{item,times[item]});times[item]++;if(it!=s.end()) s.erase(it);s.insert(node{item,times[item]});}return 0;
}

A1130(中序遍历二叉树 输出中缀表达式)

//中序遍历可以得到中缀表达式
//如果不是叶节点且不是根节点就加括号
#include<iostream>
#include<string>
using namespace std;
struct node{string v;int left,right;
}tree[21];
string ans;
int n,l,r,notroot[21]={0},root=1;
void inorder(int index){bool flag=false;if((tree[index].left!=-1||tree[index].right!=-1)&&index!=root){flag=true;ans+="(";}if(tree[index].left!=-1) inorder(tree[index].left);ans+=tree[index].v;if(tree[index].right!=-1) inorder(tree[index].right);if(flag) ans+=")";
}
int main(){cin>>n;for(int i=1;i<=n;i++){cin>>tree[i].v>>tree[i].left>>tree[i].right;if(tree[i].left!=-1) notroot[tree[i].left]=1;if(tree[i].right!=-1) notroot[tree[i].right]=1;}while(notroot[root]!=0) root++;inorder(root);cout<<ans;return 0;
}

A1131(30 DFS 、unordered_map邻接矩阵、 难题 )

#include<cstdio>
#include<vector>
#include<unordered_map>
using namespace std;
vector<vector<int> > v(10000);
int visit[10000] = { 0 }, mincnt, mintransfer, st, ed;
unordered_map<int, int> line;
vector<int> path, temppath;
int transfercnt(vector<int> a) {int cnt = -1, preline = 0;for (int i = 1; i < a.size(); i++) {if (line[a[i - 1] * 10000 + a[i]] != preline) cnt++;preline = line[a[i - 1] * 10000 + a[i]];}return cnt;
}
void dfs(int node, int cnt) {if (node == ed && (cnt < mincnt || (cnt == mincnt && transfercnt(temppath) < mintransfer))) {mincnt = cnt;mintransfer = transfercnt(temppath);path = temppath;return;}for (int i = 0; i < v[node].size(); i++) {if (visit[v[node][i]] == 0) {visit[v[node][i]] = 1;temppath.push_back(v[node][i]);dfs(v[node][i], cnt + 1);visit[v[node][i]] = 0;temppath.pop_back();}}
}
int main() {int n, m, pre, temp, k;scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d%d", &m, &pre);for (int j = 1; j < m; j++) {scanf("%d", &temp);v[pre].push_back(temp);v[temp].push_back(pre);line[pre * 10000 + temp]=line[temp*10000+pre] = i;pre = temp;}}scanf("%d", &k);while (k--) {mintransfer = mincnt = 1000000000;scanf("%d%d", &st, &ed);temppath.clear();temppath.push_back(st);visit[st] = 1;dfs(st, 0);visit[st] = 0;printf("%d\n", mincnt);int preline = 0, pretransfer = st;for (int i = 1; i < path.size(); i++) {if (line[path[i - 1] * 10000 + path[i]] != preline) {if (preline != 0) printf("Take Line#%d from %04d to %04d.\n", preline, pretransfer, path[i - 1]);preline = line[path[i - 1] * 10000 + path[i]];pretransfer = path[i - 1];}}printf("Take Line#%d from %04d to %04d.\n", preline, pretransfer, ed);}return 0;
}

A1132(20 字符串 水题)

#include<iostream>
#include<string>
using namespace std;
int main() {string s, s1, s2;int n, len;cin >> n;while (n--) {cin >> s;len = s.size();s1 = s.substr(0, len / 2);s2 = s.substr(len / 2, len / 2);if (stoi(s1) == 0 || stoi(s2) == 0) cout << "No" << endl;else if (stoi(s) % (stoi(s1) * stoi(s2)) == 0) cout << "Yes" << endl;else cout << "No" << endl;}return 0;
}

A1133(链表 重新排列)

#include<cstdio>
#include<vector>
using namespace std;
struct node {int id, data, next;
};
vector<node> v, ans;
int main() {node a[100010];int begin, n, k;int id, data, next;scanf("%d%d%d", &begin, &n, &k);for (int i = 0; i < n; i++) {scanf("%d%d%d", &id, &data, &next);a[id] = { id,data,next };}for (; begin != -1; begin = a[begin].next) v.push_back(a[begin]);for (int i = 0; i < v.size(); i++) {if (v[i].data < 0) ans.push_back(v[i]);}for (int i = 0; i < v.size(); i++) {if (v[i].data >= 0 && v[i].data <= k) ans.push_back(v[i]);}for (int i = 0; i < v.size(); i++) {if (v[i].data > k) ans.push_back(v[i]);}for (int i = 0; i < ans.size() - 1; i++) {printf("%05d %d %05d\n", ans[i].id, ans[i].data, ans[i + 1].id);}printf("%05d %d -1", ans[ans.size() - 1].id, ans[ans.size() - 1].data);return 0;
}

A1134(25 hash散列)

#include<cstdio>
#include<vector>
using namespace std;
int main(){int n,m,k,exist[10001];scanf("%d%d",&n,&m);vector<int> v[10001];for(int i=0;i<m;i++){int v1,v2;scanf("%d%d",&v1,&v2);v[i].push_back(v1);v[i].push_back(v2);} scanf("%d",&k);while(k--){int num,hashtable[10001]={0},temp,flag=1;scanf("%d",&num);for(int i=0;i<num;i++){scanf("%d",&temp);hashtable[temp]=1;}for(int i=0;i<m;i++){if(hashtable[v[i][0]]!=1&&hashtable[v[i][1]]!=1){flag=0;break;}}if(flag==1) printf("Yes\n");else printf("No\n");}return 0;
}

A1135(30 判断红黑树 递归判断)

#include<cstdio>
#include<algorithm>
using namespace std;
struct node{int v;node*left,*right;
};
void create(node*&root,int v){if(root==NULL){root=new node;root->v=v;root->left=root->right=NULL;return;}if(abs(v)<=abs(root->v)) create(root->left,v);else create(root->right,v);
}
bool judge1(node*root){if(root==NULL) return true;if(root->v<0){if(root->left!=NULL&&root->left->v<0) return false;if(root->right!=NULL&&root->right->v<0) return false;}return judge1(root->left)&&judge1(root->right);
}
int getnum(node*root){if(root==NULL) return 0;int l=getnum(root->left);int r=getnum(root->right);return root->v>0? max(l,r)+1 :max(l,r);
}
bool judge2(node*root){if(root==NULL) return true;int l=getnum(root->left);int r=getnum(root->right);if(l!=r) return false;return judge2(root->left)&&judge2(root->right);
}
int main(){int k,n,temp;scanf("%d",&k);while(k--){node*root=NULL;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&temp);create(root,temp);}if(root->v>0&&judge1(root)&&judge2(root)) printf("Yes\n");else printf("No\n");}return 0;
}

A1136 (20 回文串 字符串)

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
string revs(string s){reverse(s.begin(),s.end());return s;
}
string add(string s1,string s2){string s=s1;int carry=0;for(int i=s1.size()-1;i>=0;i--){s[i]=((s1[i]-'0')+(s2[i]-'0')+carry)%10+'0';carry=((s1[i]-'0')+(s2[i]-'0')+carry)/10;}if(carry>0) s='1'+s;return s;
}
int main(){string s,rev;cin>>s;if(s==revs(s)) {cout<<s<<" is a palindromic number.";return 0;}for(int i=1;i<=10;i++){string sum=add(s,revs(s));cout<<s<<" + "<<revs(s)<<" = "<<sum<<endl;s=sum;if(s==revs(s)){cout<<s<<" is a palindromic number.";break;}else{if(i==10) cout<<"Not found in 10 iterations.";}}return 0;
}

A1137(25 map 排序)

#include<bits/stdc++.h>
using namespace std;
struct node {string name;int gp,gm,gf,g;
};
int p,m,n,index1=1;
map<string,int> mp;
vector<node> v,ans;
bool cmp(node a,node b){return a.g!=b.g? a.g>b.g : a.name<b.name;
}
int main() {scanf("%d%d%d",&p,&m,&n);string name;int sc;for(int i=0; i<p; i++) {cin>>name>>sc;if(sc>=200){v.push_back(node {name,sc,-1,-1,0});mp[name]=index1++;}}for(int i=0;i<m;i++){cin>>name>>sc;if(mp[name]!=0) v[mp[name]-1].gm=sc;}for(int i=0;i<n;i++){cin>>name>>sc;if(mp[name]!=0){int id=mp[name]-1;v[id].gf=sc;if(v[id].gf>=v[id].gm) v[id].g=v[id].gf;if(v[id].gf<v[id].gm)    v[id].g=round(0.4*v[id].gm+0.6*v[id].gf);}}for(int i=0;i<v.size();i++){if(v[i].g>=60) ans.push_back(v[i]);}sort(ans.begin(),ans.end(),cmp);for(int i=0;i<ans.size();i++){printf("%s %d %d %d %d\n",ans[i].name.c_str(),ans[i].gp,ans[i].gm,ans[i].gf,ans[i].g);}return 0;
}

A1138(25 树 前序加中序输出后序)

#include <iostream>
#include <vector>
using namespace std;
vector<int> pre, in;
bool flag = false;
void postOrder(int prel, int inl, int inr) {if (inl > inr || flag == true) return;int i = inl;while (in[i] != pre[prel]) i++;postOrder(prel+1, inl, i-1);postOrder(prel+i-inl+1, i+1, inr);if (flag == false) {printf("%d", in[i]);flag = true;}
}
int main() {int n;scanf("%d", &n);pre.resize(n);in.resize(n);for (int i = 0; i < n; i++) scanf("%d", &pre[i]);for (int i = 0; i < n; i++) scanf("%d", &in[i]);postOrder(0, 0, n-1);return 0;
}

A1139(30 逻辑题 unordered_map)

#include<unordered_map>
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{int a,b;
};
bool cmp(const node &i,const node &j){return i.a!=j.a? i.a<j.a : i.b<j.b;
}
unordered_map<int ,bool> isfri;
int main(){int n,m,k,c,d;cin>>n>>m;string a,b;vector<int> v[10000];for(int i=0;i<m;i++){cin>>a>>b;if(a.length()==b.length()) {v[abs(stoi(a))].push_back(abs(stoi(b)));v[abs(stoi(b))].push_back(abs(stoi(a)));}
isfri[abs(stoi(a))*10000+abs(stoi(b))]=isfri[abs(stoi(b))*10000+abs(stoi(a))]=true;}cin>>k;while(k--){vector<node> ans;cin>>c>>d;for(int i=0;i<v[abs(c)].size();i++){for(int j=0;j<v[abs(d)].size();j++){if(v[abs(c)][i]==abs(d)||v[abs(d)][j]==abs(c)) continue;if(isfri[v[abs(c)][i]*10000+v[abs(d)][j]]==true) ans.push_back(node{v[abs(c)][i],v[abs(d)][j]});}}sort(ans.begin(),ans.end(),cmp);cout<<ans.size()<<endl;for(int i=0;i<ans.size();i++) printf("%04d %04d\n",ans[i].a,ans[i].b);}return 0;
}

A1140(20 字符串)

#include<iostream>
#include<string>
using namespace std;
int main(){string s;int n,j;cin>>s>>n;for(int x=1;x<n;x++){string ans;for(int i=0;i<s.length();i=j){for(j=i;j<s.length()&&s[j]==s[i];j++);ans+=s[i]+to_string(j-i);}s=ans;}cout<<s;return 0;
}

A1141(25 stl、排序)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<unordered_map>
using namespace std;
struct node{string name;double score=0;int num=0,fscore=0,rank;
}t[100010];
int index=0;
unordered_map<string,int> mp;
int change(string &s){for(int i=0;i<s.size();i++){if(isupper(s[i])) s[i]=tolower(s[i]);}if(mp.count(s)==0) mp[s]=index++;return mp[s];
}
bool cmp(node a,node b){if(a.fscore!=b.fscore) return a.fscore>b.fscore;else if(a.num!=b.num) return a.num<b.num;else return a.name<b.name;
}
int main(){int n;cin>>n;for(int i=0;i<n;i++){string id,code;int score,tid;cin>>id>>score>>code;tid=change(code);t[tid].name=code;t[tid].num++;if(id[0]=='B') t[tid].score+=score/1.5;if(id[0]=='A') t[tid].score+=score*1.0;if(id[0]=='T') t[tid].score+=score*1.5;}for(int i=0;i<index;i++) t[i].fscore=(int)(t[i].score);sort(t,t+index,cmp);cout<<index<<endl;t[0].rank=1;printf("%d %s %d %d\n",t[0].rank,t[0].name.c_str(),t[0].fscore,t[0].num);for(int i=1;i<index;i++){if(t[i].fscore==t[i-1].fscore) t[i].rank=t[i-1].rank;else t[i].rank=i+1;printf("%d %s %d %d\n",t[i].rank,t[i].name.c_str(),t[i].fscore,t[i].num);}return 0;
}

A1142(25 无向完全图 最大子图 两点相连)

#include<cstdio>
#include<vector>
using namespace std;
int main(){int nv,ne,m,k,g[210][210],v1,v2;scanf("%d%d",&nv,&ne);for(int i=0;i<ne;i++){scanf("%d%d",&v1,&v2);g[v1][v2]=g[v2][v1]=1;}scanf("%d",&m);while(m--){int isclique=1,ismax=1,hash[210]={0};scanf("%d",&k);vector<int> v(k);for(int i=0;i<k;i++) {scanf("%d",&v[i]);hash[v[i]]=1;}for(int i=0;i<k-1;i++){if(isclique==0) break;for(int j=i+1;j<k;j++){if(g[v[i]][v[j]]!=1){isclique=0;printf("Not a Clique\n");break;}}}if(isclique==0) continue;for(int i=1;i<=nv;i++){if(ismax==0) break;if(hash[i]!=1){for(int j=0;j<k;j++){if(g[i][v[j]]==1){if(j==k-1) ismax=0;}else break;}}}if(ismax==1) printf("Yes\n");else printf("Not Maximal\n");} return 0;
}

A1143(30 树 LCA 最低公共祖先)

#include<cstdio>
#include<unordered_map>
#include<vector>
using namespace std;
int main(){int m,n,u,v,ans;scanf("%d%d",&m,&n);unordered_map<int,bool>mp;vector<int> pre(n);for(int i=0;i<n;i++){scanf("%d",&pre[i]);mp[pre[i]]=true;}for(int i=0;i<m;i++){scanf("%d%d",&u,&v);for(int j=0;j<n;j++){if((pre[j]>=u&&pre[j]<=v)||(pre[j]<=u&&pre[j]>=v)){ans=pre[j];break;}}if(mp.count(u)==0&&mp.count(v)==0) printf("ERROR: %d and %d are not found.\n",u,v);else if(mp.count(u)==0&&mp.count(v)!=0) printf("ERROR: %d is not found.\n",u);else if(mp.count(v)==0&&mp.count(u)!=0) printf("ERROR: %d is not found.\n",v);else if(ans==u||ans==v) printf("%d is an ancestor of %d.\n",ans,ans==u?v:u);else printf("LCA of %d and %d is %d.\n",u,v,ans);}return 0;
}

A1145(25 hash 平方探查)

#include<cstdio>
#include<vector>
#include<cmath>
using namespace std;
bool isprime(int x){int sqr=sqrt(1.0*x);for(int i=2;i<=sqr;i++){if(x%i==0) return false;}return true;
}
int main(){int msize,n,m,a;scanf("%d%d%d",&msize,&n,&m);while(!isprime(msize)) msize++;vector<int> v(msize);for(int i=0;i<n;i++){scanf("%d",&a);int flag=0;for(int j=0;j<msize;j++){int pos=(a+j*j)%msize;if(v[pos]==0){flag=1;v[pos]=a;break;}}if(flag==0) printf("%d cannot be inserted.\n",a);}int cnt=0,temp;for(int i=0;i<m;i++){scanf("%d",&temp);for(int j=0;j<=msize;j++){//注意这里的<=，因为要回到初始的位置才知道是否能插入cnt++;if(v[(temp+j*j)%msize]==temp||v[(temp+j*j)%msize]==0) break;}}printf("%.1f",cnt*1.0/m);return 0;
}

A1144(20 找丢失的最小正数)

#include<iostream>
#include<map>
using namespace std;
int main(){int n,ans=1,temp;map<int,int> mp;cin>>n;for(int i=0;i<n;i++){cin>>temp;if(temp>0) mp[temp]++;}while(mp[ans]!=0) ans++;cout<<ans;return 0;
}

A1145(25 hash 平方探查法插入与查找)

#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
bool isprime(int n){if(n<=1) return false;int sqr=sqrt(n*1.0);for(int i=2;i<=sqr;i++){if(n%i==0) return false;}return true;
}
int main(){int msize,n,m,temp,cnt=0;;scanf("%d%d%d",&msize,&n,&m);while(!isprime(msize)) msize++;vector<int> v(msize,0);//注意初始化数组元素for(int i=0;i<n;i++){scanf("%d",&temp);int flag=0;for(int j=0;j<msize;j++){//注意是小于号 int pos=(temp+j*j)%msize;if(v[pos]==0){v[pos]=temp;flag=1;break;}}if(flag==0) printf("%d cannot be inserted.\n",temp);}for(int i=0;i<m;i++){scanf("%d",&temp);for(int j=0;j<=msize;j++){//注意这里是等号,与插入时不同，要回到初始位置才知道查找失败 int pos=(temp+j*j)%msize;cnt++;if(v[pos]==temp||v[pos]==0) break;//找到或没找到}}printf("%.1f",cnt*1.0/m);
}

A1146(25 判断是否为拓扑排序序列)

#include<cstdio>
#include<vector>
using namespace std;
int main(){int n,m,k,in[1001]={0},flag=0,temp;scanf("%d%d",&n,&m);vector<int> v[1001];for(int i=0;i<m;i++){int v1,v2;scanf("%d%d",&v1,&v2);v[v1].push_back(v2);in[v2]++;}scanf("%d",&k);for(int i=0;i<k;i++){vector<int> tin(in,in+n+1);int judge=1;for(int j=0;j<n;j++){scanf("%d",&temp);if(tin[temp]!=0) judge=0;for(int x=0;x<v[temp].size();x++) tin[v[temp][x]]--;}if(judge==1) continue;if(flag==1) printf(" ");printf("%d",i);flag=1; }return 0;
}

A1147(30 判断大顶堆小顶堆 后序遍历)

#include<cstdio>
#include<vector>
using namespace std;
int m, n, num = 0;
vector<int>v;
void dfs(int index) {//后序遍历if (index > n) return;dfs(index * 2);dfs(index * 2 + 1);printf("%d", v[index]);if (num < n - 1)printf(" ");num++;
}
int main() {scanf_s("%d%d", &m, &n);while (m--) {v.resize(n+1);bool flagmin = true, flagmax = true;for (int i = 1; i <= n; i++) scanf("%d", &v[i]);for (int i = 2; i <= n; i++) {if (v[i] >= v[i / 2]) flagmax = false;if (v[i] <= v[i / 2]) flagmin = false;}if (flagmin) printf("Min Heap\n");else if (flagmax) printf("Max Heap\n");else if (flagmax == false && flagmin == false) printf("Not Heap\n");dfs(1);num = 0;printf("\n");}return 0;
}

A1148(狼人杀 找到两个狼人)

#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
int main(){int n;cin>>n;vector<int> v(n+1);for(int i=1;i<=n;i++) cin>>v[i];for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){vector<int> lie,a(n+1,1);a[i]=a[j]=-1;for(int k=1;k<=n;k++){if(v[k]*a[abs(v[k])]<0) lie.push_back(k);}if(lie.size()==2&&a[lie[0]]+a[lie[1]]==0){cout<<i<<" "<<j;return 0;}}}cout<<"No Solution";return 0;
}

A1149(25 map)

#include<cstdio>
#include<map>
#include<vector>
using namespace std;
int main(){int n,m,g1,g2,num;scanf("%d%d",&n,&m);map<int,vector<int> > mp;while(n--){scanf("%d%d",&g1,&g2);mp[g1].push_back(g2);mp[g2].push_back(g1);}while(m--){scanf("%d",&num);vector<int> v(num);int exist[100005]={0},flag=1;for(int i=0;i<num;i++){scanf("%d",&v[i]);exist[v[i]]=1;}for(int i=0;i<num;i++){for(int j=0;j<mp[v[i]].size();j++){if(exist[mp[v[i]][j]]==1){printf("No\n");flag=0;break;}}if(flag==0) break;}if(flag==1) printf("Yes\n");}return 0;
}

A1150(25 判断循环图 输出最小路径)

#include<cstdio>
#include<map>
#include<set>
#include<vector>
using namespace std;
int main(){int n,m,k,dis[210][210]={0},mindis=0x7fffffff,ansid;scanf("%d%d",&n,&m);while(m--){int c1,c2,d;scanf("%d%d%d",&c1,&c2,&d);dis[c1][c2]=dis[c2][c1]=d;}scanf("%d",&k);for(int x=1;x<=k;x++){int num,total=0,flag=1;scanf("%d",&num);set<int> s;vector<int> v(num);for(int i=0;i<num;i++){scanf("%d",&v[i]);s.insert(v[i]);}for(int i=0;i<num-1;i++){if(dis[v[i]][v[i+1]]==0){flag=0;break;}total+=dis[v[i]][v[i+1]];} printf("Path %d:",x);if(flag==0) printf(" NA ");else printf(" %d ",total);if(v[0]!=v[num-1]||s.size()!=n||flag==0) printf("(Not a TS cycle)\n");else if(num-1==n&&s.size()==n&&v[0]==v[num-1]){if(total<mindis) {mindis=total;ansid=x;}printf("(TS simple cycle)\n");} else if(num-1!=n&&s.size()==n&&v[0]==v[num-1]){if(total<mindis) {mindis=total;ansid=x;}printf("(TS cycle)\n");} }printf("Shortest Dist(%d) = %d",ansid,mindis);return 0;
}

A1151(30 树 中序加先序求LCA)

方法一:
#include<cstdio>
#include<map>
#include<vector>
using namespace std;
const int maxn=10010;
map<int,int> pos;
vector<int> ins,pre;
void lca(int inl,int inr,int preroot,int a,int b){if(inl>inr) return;int inroot=pos[pre[preroot]],ain=pos[a],bin=pos[b];if(ain<inroot&&bin<inroot) lca(inl,inroot-1,preroot+1,a,b);else if((ain<inroot&&bin>inroot)||(ain>inroot&&bin<inroot)){printf("LCA of %d and %d is %d.\n",a,b,ins[inroot]);}else if(ain>inroot&&bin>inroot) lca(inroot+1,inr,preroot+1+(inroot-inl),a,b);else if(ain==inroot) printf("%d is an ancestor of %d.\n",a,b);else if(bin==inroot) printf("%d is an ancestor of %d.\n",b,a);
}
int main(){int m,n,a,b;scanf("%d%d",&m,&n);ins.resize(n + 1), pre.resize(n + 1);for(int i=1;i<=n;i++){scanf("%d",&ins[i]);pos[ins[i]]=i;}for(int i=1;i<=n;i++) scanf("%d",&pre[i]);for(int i=0;i<m;i++){scanf("%d%d",&a,&b);if(pos.count(a)==0&&pos.count(b)!=0) printf("ERROR: %d is not found.\n",a);if(pos.count(b)==0&&pos.count(a)!=0) printf("ERROR: %d is not found.\n",b);if(pos.count(a)==0&&pos.count(b)==0) printf("ERROR: %d and %d are not found.\n",a,b);if(pos.count(a)!=0&&pos.count(b)!=0) lca(1,n,1,a,b);}return 0;
}方法二:建树加dfs
#include<cstdio>
#include<unordered_map>
using namespace std;
const int maxn=10010;
int in[maxn],pre[maxn];
struct node{int v;node*left,*right;
};
node* create(int prel,int prer,int inl,int inr){if(prel>prer) return NULL;node*root=new node;root->v=pre[prel];int k=inl;while(in[k]!=root->v&&k<=inr) k++;int leftnum=k-inl;root->left=create(prel+1,prel+leftnum,inl,k-1);root->right=create(prel+leftnum+1,prer,k+1,inr);return root;
}
node* lca(node*root,int a,int b){if(root==NULL||root->v==a||root->v==b) return root;node*l=lca(root->left,a,b);node*r=lca(root->right,a,b);if(l==NULL) return r;if(r==NULL) return l;return root;
}
int main(){int m,n,u,v,res;scanf("%d%d",&m,&n);unordered_map<int,bool>mp;for(int i=0;i<n;i++){scanf("%d",&in[i]);mp[in[i]]=true;}for(int i=0;i<n;i++) scanf("%d",&pre[i]);node*root=create(0,n-1,0,n-1);for(int i=0;i<m;i++){scanf("%d%d",&u,&v);if(mp.count(u)==0&&mp.count(v)==0) printf("ERROR: %d and %d are not found.\n",u,v);else if(mp.count(u)==0&&mp.count(v)!=0) printf("ERROR: %d is not found.\n",u);else if(mp.count(v)==0&&mp.count(u)!=0) printf("ERROR: %d is not found.\n",v);else{node*ans=lca(root,u,v);res=ans->v;if(res==u||res==v) printf("%d is an ancestor of %d.\n",res,res==u?v:u);else printf("LCA of %d and %d is %d.\n",u,v,res);}}return 0;
}

A1152(20 字符串中找第一个k位素数)

#include<iostream>
#include<string>
#include<cmath>
using namespace std;
bool isprime(string s){if(stoi(s)<=1) return false;int sqr=sqrt(stoi(s));for(int i=2;i<=sqr;i++){if(stoi(s)%i==0) return false;}return true;
}
int main(){int l,k;string s;cin>>l>>k>>s;for(int i=0;i<s.size();i++){if(i+k<=s.size()){string t=s.substr(i,k);if(isprime(t)){cout<<t;return 0;}}}cout<<"404";return 0;
}

A1153(25 模拟 排序引用传参 unordered_map)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<unordered_map>
using namespace std;
struct node {string info;int score;
};
bool cmp(const node& a, const node& b) {return a.score != b.score ? a.score > b.score : a.info < b.info;
}
int main() {int n, m, type;string t;cin >> n >> m;vector<node> v(n);for (int i = 0; i < n; i++) cin >> v[i].info >> v[i].score;for (int i = 1; i <= m; i++) {vector<node> ans;cin >> type >> t;printf("Case %d: %d %s\n", i, type, t.c_str());int sum = 0, cnt = 0;if (type == 1) {for (int j = 0; j < n; j++) {if (t[0] == v[j].info[0]) ans.push_back(v[j]);}}else if (type == 2) {for (int j = 0; j < n; j++) {if (v[j].info.substr(1, 3) == t) {cnt++;sum += v[j].score;}}if (cnt != 0) printf("%d %d\n", cnt, sum);}else {unordered_map<string, int> m;for (int j = 0; j < n; j++) {if (v[j].info.substr(4, 6) == t) m[v[j].info.substr(1, 3)]++;}for (auto it : m) ans.push_back({ it.first,it.second });}sort(ans.begin(), ans.end(), cmp);for (int j = 0; j < ans.size(); j++) printf("%s %d\n", ans[j].info.c_str(), ans[j].score);if (((type == 1 || type == 3) && ans.size() == 0) || (type == 2 && cnt == 0)) printf("NA\n");}return 0;
}

A1154(25 图 边的两端点的判断 )

#include<cstdio>
#include<vector>
#include<set>
using namespace std;
struct node{int v1,v2;
};
int main(){int n,m,k;scanf("%d%d",&n,&m);vector<node> v(m);vector<int> colo(n);for(int i=0;i<m;i++) scanf("%d%d",&v[i].v1,&v[i].v2);scanf("%d",&k);while(k--){int flag=1;set<int> s;for(int i=0;i<n;i++){scanf("%d",&colo[i]);s.insert(colo[i]);}for(int i=0;i<m;i++){if(colo[v[i].v1]==colo[v[i].v2]){flag=0;break;}}if(flag==0) printf("No\n");else printf("%d-coloring\n",s.size());}return 0;
}

A1155(30 完全二叉树 判断大顶堆小顶堆 dfs 打印路径)

#include<bits/stdc++.h>
using namespace std;
int n,tree[1005];
vector<int> temp;
void dfs(int index){if(index*2>n&&index*2+1>n){temp.push_back(index);for(int i=0;i<temp.size();i++){if(i>0) printf(" ");printf("%d",tree[temp[i]]);}printf("\n");temp.pop_back();return;}temp.push_back(index);if(index*2+1<=n) dfs(index*2+1);if(index*2<=n) dfs(index*2);temp.pop_back();
}
int main(){cin>>n;bool minheap=true,maxheap=true;for(int i=1;i<=n;i++) cin>>tree[i];for(int i=n;i>=2;i--){if(tree[i]>tree[i/2]) maxheap=false;if(tree[i]<tree[i/2]) minheap=false;}dfs(1);if(minheap) printf("Min Heap\n");else if(maxheap) printf("Max Heap\n");else printf("Not Heap\n");return 0;
}

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