PAT

PAT (Advanced Level) Practice - II（1051-1100）

--------------------------------------------------------------------------------

1001-1167（已更新至最新！）

1051-1100

链接：https://pan.baidu.com/s/1ZKiso4L947R3MZj7sxVtaw
提取码：sp2z
--------------------------------------------------------------------------------

题目传送门

PAT (Advanced Level) Practice
--------------------------------------------------------------------------------

序

这里是PAT (Advanced Level) Practice - II（1051-1100）
这是一个全新的起点，我们完成了乙级考试的小目标，向着甲级考试的大目标前进，你将会看到一片新的世界，在数据结构和算法的大海里遨游
首先，我们讲讲两种考试的区别：
1.题面描述不同，乙级考试采用中文描述，甲级考试采用英文描述
2.题目数量不同，乙级考试是5道题，甲级考试是4道题
3.题目的分值不同，乙级考试的分值分布为15/20/20/20/25，甲级考试的分值分布为20/25/25/30
4.难度不同，甲级考试相对乙级考试来说会增加一个难度级别，主要体现在对数据结构的掌握上，如树和图——主要有BST/BBST/CBT/Heap/DAG,算法方面在原有的难度基础之上加深一点（比方说大模拟），引入了一些简单DP
知己知彼方能百战不殆，我们了解基本的区别之后，大家也不用太担心考试的难度问题，毕竟数据规模n不像竞赛要求那么高，很多算法用最普通的枚举方法往往也能取得满分！
接下来就是题解部分了，每道算法题都标注有对应的算法标签，对于那些易错、较难或是测试点比较特殊的题目会着重标注，本章推荐的题目有：

1053 Path of Equal Weight (30 分) 1079 Total Sales of Supply Chain (25 分) 1086 Tree Traversals Again (25 分) 1090 Highest Price in Supply Chain (25 分)	树，二叉树 + DFS
1056 Mice and Rice (25 分)	大模拟
1057 Stack (30 分)	分块/ST表
1067 Sort with Swap(0, i) (25 分)	数论 + 模拟
1072 Gas Station (30 分) 1087 All Roads Lead to Rome (30 分)	图论 + Dijkstra

--------------------------------------------------------------------------------

1051 Pop Sequence (25 分)

算法标签: 模拟 + 栈
注意点: 本题模拟栈混洗的过程，并要求判断混洗后的结果是否为出栈序列，因此，对于每行输入，我们都需要开辟一个新的栈，不断压入元素，若该元素为出栈序列的首部，则弹出，检验下一个出栈首部，所有数据全部输入完成，且模拟过程结束后栈为空，表示为合法的出栈序列，否则就是不合法的序列

#include<bits/stdc++.h>
using namespace std;
int a[1005];
int M,N,K;
void check(){stack<int> s;bool f = true;int now = 1;for(int i=1;i<=N;i++){s.push(i);if(s.size()>M){f = false;break;}while(!s.empty() && s.top()==a[now]){s.pop();now++;}}if(!s.empty()){f = false;}if(f){printf("YES\n");}else{printf("NO\n");}
}int main(){cin >> M >> N >> K;for(int t=1;t<=K;t++){for(int i=1;i<=N;i++){scanf("%d",&a[i]);}check();} return 0;
}

1052 Linked List Sorting (25 分)

算法标签: 排序 + 链表

#include<bits/stdc++.h>
using namespace std;
struct node{int value;int next;
};
const int maxn = 1e5+5;
node n[maxn];
struct out{int add;int value;
};
out o[maxn];
bool cmp(out o1,out o2){return o1.value < o2.value;
}
int main(){int N;cin >> N;int start;cin >> start;int cnt = 0;for(int i=0;i<N;i++){int t;scanf("%d",&t); scanf("%d%d",&n[t].value,&n[t].next);}while(start!=-1){o[cnt].add = start;o[cnt].value = n[start].value;cnt++;start = n[start].next;}sort(o,o+cnt,cmp);if(cnt==0){printf("0 -1\n");}else{printf("%d %05d\n",cnt,o[0].add);}for(int i=0;i<cnt;i++){if(i==cnt-1){printf("%05d %d -1\n",o[i].add,o[i].value);}else{printf("%05d %d %05d\n",o[i].add,o[i].value,o[i+1].add);}}return 0;
}

1053 Path of Equal Weight (30 分)

算法标签: 树，二叉树 + DFS
注意点: 一道好题！
这道题考察了树的DFS过程，从树根结点开始，遍历所有的元素结点，并统计PWL（带权路径长度）：
1.若当前的PWL>S，则直接返回（剪枝）
2.若PWL==s，且当前访问的结点是叶结点，那么保存路径至vector中
所有路径记录完毕后，记得对所有路径进行排序，这个步骤很重要，因为不排序无法通过加强测试的数据点！

#include<bits/stdc++.h>
using namespace std;
int N,M,S;struct node{int weight;vector<int> child;
};
node n[105];
int path[105];bool cmp(int id1,int id2){return n[id1].weight > n[id2].weight;
}
vector<int> a[105];
int cnt = 0;bool cmp2(vector<int> a,vector<int> b){return a>b;
}void dfs(int depth,int index,int sum){if(sum>S){return;}else if(sum==S){if(n[index].child.size()==0){for(int i=0;i<depth;i++){a[cnt].push_back(path[i]);}cnt++;return;}else{return;}}else{for(int i=0;i<n[index].child.size();i++){int temp = n[index].child[i];path[depth] = n[temp].weight;dfs(depth+1,temp,sum+n[temp].weight);}}return;
}
int main() {cin >> N >> M >> S;for(int i=0;i<N;i++){cin >> n[i].weight;}for(int i=0;i<M;i++){int id;cin >> id;int k;cin >> k;for(int j=0;j<k;j++){int t;cin >> t;n[id].child.push_back(t);}sort(n[id].child.begin(),n[id].child.end(),cmp); }path[0] = n[0].weight;dfs(1,0,n[0].weight);sort(a,a+cnt,cmp2);for(int i=0;i<cnt;i++){int l = a[i].size();for(int j=0;j<l;j++){if(j==0){printf("%d",a[i][j]);}else{printf(" %d",a[i][j]);}}printf("\n");}return 0;
}

1054 The Dominant Color (20 分)

算法标签: 数论
注意点: The Dominant Color实质是求像素点的众数！

#include<bits/stdc++.h>
using namespace std;
map<int,int> mp;
int main(){int M,N;cin >> M >> N;for(int i=0;i<N;i++){for(int j=0;j<M;j++){int t;cin >> t; if(mp.find(t)!=mp.end()){mp[t]++;}else{mp[t] = 1;}}}int maxn = -1;int max_num = 0;for(auto it=mp.begin();it!=mp.end();it++){if(it->second>maxn){maxn = it->second;max_num = it->first;}}cout << max_num << endl;return 0;
}

1055 The World’s Richest (25 分)

算法标签: 排序

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;struct people{char name[10];int age;int worth;
};people p[maxn];bool cmp1(people p1,people p2){if(p1.worth!=p2.worth){return p1.worth > p2.worth;}else if(p1.age!=p2.age){return p1.age < p2.age;}else{return strcmp(p1.name,p2.name)<0;}
}
people p_valid[maxn];int main(){cin.tie(0);cout.tie(0);int N,K;scanf("%d%d",&N,&K);for(int i=0;i<N;i++){scanf("%s%d%d",p[i].name,&p[i].age,&p[i].worth);}sort(p,p+N,cmp1);int cnt = 0;int t_age = p[0].age;int count = 0;for(int i=0;i<N;i++){if(t_age == p[i].age && count<100){strcpy(p_valid[cnt].name,p[i].name);p_valid[cnt].age = p[i].age;p_valid[cnt].worth = p[i].worth;count++;cnt++;}else{t_age = p[i].age;strcpy(p_valid[cnt].name,p[i].name);p_valid[cnt].age = p[i].age;p_valid[cnt].worth = p[i].worth;cnt ++;count = 1;}}for(int i=0;i<K;i++){int M,s_age,e_age;bool f = false;scanf("%d%d%d",&M,&s_age,&e_age);int count = 0;cout << "Case #" << i+1 << ":" << endl;for(int j=0;j<cnt;j++){if(p_valid[j].age >= s_age && p_valid[j].age <= e_age && count<M){printf("%s %d %d\n",p_valid[j].name,p_valid[j].age,p_valid[j].worth);f = true;count++;}}if(f){continue;}else{cout << "None" << endl;}}return 0;
}

1056 Mice and Rice (25 分)

算法标签: 大模拟
注意点: 一道好题！
本题模拟分组比赛的过程，这里有个小技巧，可以边模拟边计算出老鼠们的排名，我们假设晋级下一轮的老鼠为x只，那么x等价于分组的数量group，因此，这一轮淘汰的老鼠分列第group+1名。
那么如何求group呢？其实很简单，本轮比赛老鼠的数量除以组内老鼠的数量 g r o u p = N N G group=\frac{N}{NG} group=NGN即可求得，下一轮老鼠的总数就是group(即group代替了N)

#include<bits/stdc++.h>
using namespace std;
int NP,NG;
int order[1005];struct mouse{int rank;int weight;
};
mouse m[1005];queue<int> q;int main(){cin >> NP >> NG;for(int i=0;i<NP;i++){cin >> m[i].weight;}  for(int i=0;i<NP;i++){cin >> order[i];}for(int i=0;i<NP;i++){q.push(order[i]);}int temp = NP;int 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=0;j<NG;j++){if(i*NG+j>=temp){break;}int t = q.front();if(m[t].weight>m[k].weight){k = t;}m[t].rank = group+1;q.pop();}q.push(k);}temp = group;}int k = q.front();m[k].rank = 1;for(int i=0;i<NP;i++){if(i==NP-1){printf("%d",m[i].rank);}else{printf("%d ",m[i].rank);}}return 0;
}

1057 Stack (30 分)

算法标签: 分块/ST表
注意点: 一道好题！
这道题如果按照普通枚举的方法肯定是TLE的，对于若干次查询，我们如何做到快速的找到中位数呢？
有2个方法，一个是中位数算法，另一个是分块法，这里不讲中位数算法，有兴趣可查阅相关的资料，接下来主要讲讲分块算法
分块，就是把一个哈希表分成若干块，以块为单位的形式查找，那么这里我们将整个区间分成 O ( n ) O(\sqrt n) O(n )块，每块中含有 O ( n ) O(\sqrt n) O(n )个元素，每次查询先查询块，然后再查询块内的元素，因此可在 O ( n ) O(\sqrt n) O(n )的时间内找到中位数并输出，能够满足题目的查询速度要求

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
const int sq = (int)sqrt(maxn);
int N;
int block[350];
int a[maxn];
stack<int> st;
int main(){scanf("%d",&N);for(int i=0;i<N;i++){string s;cin >> s;if(s=="Pop"){if(st.empty()){printf("Invalid\n");}else{int t = st.top();printf("%d\n",t);st.pop();a[t]--;block[t/sq]--;}}else if(s=="PeekMedian"){if(st.empty()){printf("Invalid\n");}else{int t = st.size();int K; if(t%2){K = (t+1)/2;}else{K = t/2;}int cnt1 = 0;int sum = 0;while(sum+block[cnt1]<K){sum += block[cnt1];cnt1++;}int cnt2 = sq*cnt1;while(sum<K){sum += a[cnt2];cnt2++;}printf("%d\n",cnt2-1);  }}else if(s=="Push"){int t;scanf("%d",&t);st.push(t);block[t/sq]++;a[t]++;}}return 0;
}

1058 A+B in Hogwarts (20 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;
string str1,str2;
int a[3];
int b[3];
int c[3];int main(){cin >> str1 >> str2;int site1 = str1.find('.');int site2 = str1.find('.',site1+1);for(int i=0;i<site1;i++){a[0] = a[0]*10 + str1[i] - '0'; } for(int i=site1+1;i<site2;i++){a[1] = a[1]*10 + str1[i] - '0';}for(int i=site2+1;i<str1.size();i++){a[2] = a[2]*10 + str1[i] - '0';} site1 = str2.find('.');site2 = str2.find('.',site1+1);for(int i=0;i<site1;i++){b[0] = b[0]*10 + str2[i] - '0'; } for(int i=site1+1;i<site2;i++){b[1] = b[1]*10 + str2[i] - '0';}for(int i=site2+1;i<str2.size();i++){b[2] = b[2]*10 + str2[i] - '0';} int jw = 0;for(int i=2;i>=0;i--){a[i] = a[i] + b[i];if(i==2){ a[i] = a[i] + jw; jw = a[i]/29;a[i] = a[i] % 29;}else if(i==1){a[i] = a[i] + jw;jw = a[i]/17;a[i] = a[i] % 17;}else{a[i] += jw;}}cout << a[0] << "." << a[1] << "." << a[2] << endl; return 0;
}

1059 Prime Factors (25 分)

算法标签: 数论 + 素数筛

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
bool a[maxn] = {1,1,0};
int b[maxn];
int cnt = 0;
void init(){for(int i=2;i<maxn;i++){if(!a[i]){for(int j=2*i;j<maxn;j+=i){a[j] =1;}}}for(int i=2;i<maxn;i++){if(!a[i]){b[cnt++] = i;}}
}struct factor{int p;int e = 0;
};
factor f[maxn];
int index_t = 0;
int main(){init();int N;scanf("%d",&N);if(N==1){printf("1=1");return 0;}int out = N;for(int i=0;i<cnt;i++){if(N%b[i]==0){f[index_t].p = b[i];while(N%b[i]==0 && N!=1){N/=b[i];f[index_t].e++;}index_t++;}if(N==1){break;}}if(N>1){f[index_t].p = N;f[index_t].e++;index_t++;}printf("%d=",out);for(int i=0;i<index_t;i++){if(i==0){if(f[i].e == 1){printf("%d",f[i].p);}else{printf("%d^%d",f[i].p,f[i].e);}}else{printf("*");if(f[i].e == 1){printf("%d",f[i].p);}else{printf("%d^%d",f[i].p,f[i].e);}}}return 0;
}

1060 Are They Equal (25 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;int main(){int N;scanf("%d",&N);string s1,s2;cin >> s1 >> s2;while(1){if(s1[0]=='0'){s1.erase(s1.begin());}else{break;}}while(1){if(s2[0]=='0'){s2.erase(s2.begin());}else{break;}}int len1 = s1.size();int len2 = s2.size();int E1 = 0;int E2 = 0;string r1 ="";string r2 ="";if(s1[0]=='.'){for(int i=1;i<len1;i++){if(s1[i]=='0'){E1--;}else{break;}}  int t = N;int now = 1 + abs(E1);while(t--){if(now<len1){r1 += s1[now];now++;}else{r1 += "0";}}}else{for(int i=0;i<len1;i++){if(s1[i]=='.'){break;}else{E1++;}}int now = 0;int t=N;while(t){if(now<len1){if(s1[now]!='.'){r1 += s1[now];t--;}now++;}else{r1 += "0";t--;}}}if(s2[0]=='.'){for(int i=1;i<len2;i++){if(s2[i]=='0'){E2--;}else{break;}}  int t = N;int now = 1 + abs(E2);while(t--){if(now<len2){r2 += s2[now];now++;}else{r2 += "0";}}}else{for(int i=0;i<len2;i++){if(s2[i]=='.'){break;}else{E2++;}}int t = N;int now = 0;while(t){if(now<len2){if(s2[now]!='.'){r2 += s2[now];t--;}now++;}else{r2 += "0";t--;}}}if(E1==E2 && r1==r2){printf("YES 0.%s*10^%d\n",r1.c_str(),E1);}else if(r1==r2 && r1[0]=='0' && r2[0]=='0'){printf("YES 0.%s*10^%d\n",r1.c_str(),0);}else{printf("NO 0.%s*10^%d 0.%s*10^%d\n",r1.c_str(),E1,r2.c_str(),E2);}return 0;
}

1061 Dating (20 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;
string str[7] = {"MON","TUE","WED","THU","FRI","SAT","SUN"};
int isupper(char c){if(c>='A' && c<='G'){return 1;}else{return 0;}
}int isalpha(char c){if(c>='A' && c<='Z'){return 1;}else if(c>='a' && c<='z'){return 1;}else{return 0;}
}int main(){string s1,s2,s3,s4;char day,hour;int minute;cin >> s1 >> s2 >> s3 >> s4;int size1 = min(s1.size(),s2.size());int size2 = min(s3.size(),s4.size());int count = 0;for(int i=0;i<size1;i++){if(s1[i]==s2[i] && count == 0 && isupper(s1[i]) && isupper(s2[i])){day = s1[i];count++;}else if(s1[i]==s2[i] && ((s1[i]>='0' && s1[i]<='9')||(s2[i]>='A' && s2[i]<='N')) && count!=0){hour = s1[i];break;}}    for(int i=0;i<size2;i++){if(s3[i]==s4[i] && isalpha(s3[i]) && isalpha(s4[i])){minute = i; break;} }string daystr = str[(int)(day-'A')];int h;if(isalpha(hour)){h = hour - 'A' + 10;}else{h = hour -'0';}cout << daystr << " ";if(h<10){cout << "0" << h;}else{cout << h ;}cout << ":";if(minute<10){cout << "0" << minute;}else{cout << minute;}cout << endl;return 0;
}

1062 Talent and Virtue (25 分)

算法标签: 排序

#include<bits/stdc++.h>
using namespace std;
int N,L,H;
typedef struct score{int IDnum;int Dscore;int Cscore;
}score;
const int maxn = 1e5+5;
score DC1[maxn];
score DC2[maxn];
score DC3[maxn];
score DC4[maxn];
bool cmp(score s1,score s2){if(s1.Cscore+s1.Dscore>s2.Cscore+s2.Dscore){return true;}else if(s1.Cscore+s1.Dscore == s2.Cscore+s2.Dscore){if(s1.Dscore>s2.Dscore){return true;}else if (s1.Dscore == s2.Dscore){if(s1.IDnum<s2.IDnum){return true;}else{return false;}}else{return false;}}else{return false;}
}int main(){cin >> N >> L >> H;int first = 0,second =0,third =0,fourth =0;int nopass = 0;for(int i=0;i<N;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);if(b<L || c<L){nopass++;continue;}else if(b>=H && c>=H){DC1[first].IDnum = a;DC1[first].Dscore = b;DC1[first++].Cscore = c;}else if(b>=H && c<H){DC2[second].IDnum = a;DC2[second].Dscore = b;DC2[second++].Cscore = c;}else if(b<H && c<H && b>=c){DC3[third].IDnum = a;DC3[third].Dscore = b;DC3[third++].Cscore = c;}else{DC4[fourth].IDnum = a;DC4[fourth].Dscore = b;DC4[fourth++].Cscore = c;}}cout << N-nopass << endl;sort(DC1,DC1+first,cmp);sort(DC2,DC2+second,cmp);sort(DC3,DC3+third,cmp);sort(DC4,DC4+fourth,cmp);for(int i=0;i<first;i++){printf("%d %d %d\n",DC1[i].IDnum,DC1[i].Dscore,DC1[i].Cscore);}for(int i=0;i<second;i++){printf("%d %d %d\n",DC2[i].IDnum,DC2[i].Dscore,DC2[i].Cscore);}for(int i=0;i<third;i++){printf("%d %d %d\n",DC3[i].IDnum,DC3[i].Dscore,DC3[i].Cscore);}for(int i=0;i<fourth;i++){printf("%d %d %d\n",DC4[i].IDnum,DC4[i].Dscore,DC4[i].Cscore);}return 0;
}

1063 Set Similarity (25 分)

算法标签: 集合
注意点: 本题考察的重点是集合，由于集合中的元素不能重复，因此，可用STL的set进行操作，每次查询输入两个集合的编号，我们采用枚举法判断集合的相似度，最后保留1位小数后输出

#include<bits/stdc++.h>
using namespace std;
set<int> a[55];
double query[60][60];
int main(){int N;scanf("%d",&N);for(int i=1;i<=N;i++){int M;scanf("%d",&M);for(int j=0;j<M;j++){int t;scanf("%d",&t);a[i].insert(t);}}int K;scanf("%d",&K);for(int i=0;i<K;i++){int g,h;scanf("%d%d",&g,&h);set<int>::iterator it1;set<int>::iterator it2;int Nc = 0;for(it1 = a[g].begin(),it2 = a[h].begin();it1!=a[g].end() && it2!=a[h].end();){if(*it1 == *it2){Nc++;it1++;it2++;}else if(*it1<*it2){it1++;}else{it2++;}}int Nn = a[g].size() + a[h].size() - Nc;double r = (Nc * 1.0) /(Nn * 1.0);printf("%.1lf%%\n",r*100.0);}return 0;
}

1064 Complete Binary Search Tree (30 分)

算法标签: 树，二叉树 + CBT + BST
注意点: 本题不必模拟树中结点的插入过程，首先输入数据后，对数据进行排序，由于BST的中序遍历必定是有序的，因此中序遍历的结果就是排序后的数组，那么层序遍历的求解就可以转化为根结点的中序遍历过程，root位置代表了当前结点的层次序号，而第k个被遍历的元素就是排序后数组的第k项

#include<bits/stdc++.h>
using namespace std;
int a[1005];
int b[1005];
int cnt = 1;
int N;void InOrder(int root){if(root>N){return;}InOrder(2*root);b[root] = a[cnt++];InOrder(2*root+1);
}int main(){cin >> N;for(int i=1;i<=N;i++){cin >> a[i];}sort(a+1,a+N+1); InOrder(1);for(int i=1;i<=N;i++){if(i==1){printf("%d",b[i]);}else{printf(" %d",b[i]);}}printf("\n");return 0;
}

1065 A+B and C (64bit) (20 分)

算法标签: 数论
注意点: 需要特判上溢和下溢的情况
上溢是指两个正数相加，结果却为负数
下溢是指两个负数相加，结果却为正数

#include<bits/stdc++.h>
using namespace std;int main(){int T;cin >> T;long long a,b,c;for(int i=1;i<=T;i++){scanf("%lld%lld%lld",&a,&b,&c);cout << "Case #" << i << ": ";long long res = a+b;if(a>0 && b>0 && res<0){cout << "true" << endl;}else if(a>0 && b>0 && res==0){cout << "true" << endl;}else if(a<0 && b<0 && res>0){cout << "false" << endl;}else if(a<0 && b<0 && res==0){cout << "false" << endl;}else if(res>c){cout << "true" << endl;}else{cout << "false" << endl;}}return 0;
}

1066 Root of AVL Tree (25 分)

算法标签: 树，二叉树 + 平衡树
注意点: 平衡树模板题，先建树，然后输出根结点的值

#include<bits/stdc++.h>
using namespace std;
int a[25];
struct node{int data;int height;node* lchild;node* rchild;
};
node* root = NULL;
node* newnode(int v){node* Node = new node;Node->data = v;Node->height = 1;Node->lchild = NULL;Node->rchild = NULL;return Node;
}
int getheight(node *root){if(root==NULL){return 0;}return root->height;
}
void updateheight(node *root){root->height = max(getheight(root->lchild),getheight(root->rchild))+1;
}
int bf(node* root){return getheight(root->lchild) - getheight(root->rchild);
}
void R(node* &root){node* temp = root->lchild;root->lchild = temp->rchild;temp->rchild = root;updateheight(root);updateheight(temp);root = temp;
}
void L(node* &root){node* temp = root->rchild;root->rchild = temp->lchild;temp->lchild = root;updateheight(root);updateheight(temp);root = temp;
}void insert(node* &root,int x){if(root==NULL){root = newnode(x);return;}if(x<root->data){insert(root->lchild,x);updateheight(root);if(bf(root)==2){if(bf(root->lchild)==1){R(root);}else{L(root->lchild);R(root);}}}else{insert(root->rchild,x);updateheight(root);if(bf(root)==-2){if(bf(root->rchild)==-1){L(root);}else{R(root->rchild);L(root);}}}
}
int main(){int N;cin >> N;for(int i=0;i<N;i++){cin >> a[i];}for(int i=0;i<N;i++){insert(root,a[i]);}cout << root->data << endl;return 0;
}

1067 Sort with Swap(0, i) (25 分)

算法标签: 数论 + 模拟
注意点: 一道好题！这道题考察了如何求解交换的最小次序，首先我们观察任意一个数列，这个数列有序当且仅当元素都在对应的位置上，即 a [ i ] = i a[i]=i a[i]=i，若要执行Swap变换，那么就必须和0进行交换，而0这个元素有序当且仅当最后一个元素有序后，0自动有序。那么该问题求解的过程就转化为除了0之外的所有元素有序时最小次数。此时问题分成2个部分：若0不在a[0]上，可交换a[a[0]]与a[0]的值，那么a[0]就位于对应位置上了；若0在a[0]上，此时我们选择从前向后未有序的第一个元素，和0交换，然后再反复执行上述操作，直到全部有序为止，输出交换的次数即为答案。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
int a[maxn];int main(){int N;scanf("%d",&N);int cnt = 0;int swapcount = 0;for(int i=0;i<N;i++){scanf("%d",&a[i]);if(a[i]!=0 && i!=a[i]){cnt++;}}int k = 1;while(cnt>0){if(a[0] == 0){for(int i=k;i<N;i++){if(i!=a[i]){k = i;break;}}swap(a[0],a[k]);swapcount++;}while(a[0]!=0){swap(a[a[0]],a[0]);swapcount++;cnt--;}   }printf("%d\n",swapcount);return 0;
}

1068 Find More Coins (30 分)

算法标签: DP + DFS
注意点: 要找出一些数字和为M，这是一道经典的0-1背包问题，建立状态转移方程:
d p [ j ] = m a x ( d p [ j ] , d p [ j − a [ i ] ] + a [ i ] ) dp[j] = max(dp[j],dp[j-a[i]]+a[i]) dp[j]=max(dp[j],dp[j−a[i]]+a[i])
由于题目中要输出字典序最小的，因此需要对数据元素倒排，然后进行DP操作，DP过程中记录路径结点，choice[i][j]表示选了第i个数后和为j，最后再逆序DFS求出每个选入的元素，依次输出即可

#include<bits/stdc++.h>
using namespace std;
int N,M;
int a[10005];
int dp[105];
bool choice[10005][105];
int ans[105];
int cnt = 0;int main(){scanf("%d%d",&N,&M);for(int i=1;i<=N;i++){scanf("%d",&a[i]); }sort(a+1,a+N+1);reverse(a+1,a+N+1);for(int i=1;i<=N;i++){for(int j=M;j>=a[i];j--){if(dp[j-a[i]]+a[i]>=dp[j]){dp[j] = max(dp[j],dp[j-a[i]]+a[i]);choice[i][j] = true;}}}if(dp[M]!=M){printf("No Solution\n");}else{int j = M;int i = N;while(j){if(choice[i][j]){ans[cnt++] = a[i];j -= a[i]; }i--;}for(int i=0;i<cnt;i++){if(i==0){printf("%d",ans[i]);}else{printf(" %d",ans[i]);}}printf("\n");}return 0;
}

1069 The Black Hole of Numbers (20 分)

算法标签: 数论 + 模拟

#include<bits/stdc++.h>
using namespace std;
int N;
int a[4];
int b[4];
int c[4];
bool cmp(int a,int b){return a>b;
}
int main(){cin >> N;a[0] = N / 1000;a[1] = (N / 100) % 10;a[2] = (N / 10) % 10;a[3] = N % 10;if((a[0]==a[1])&&(a[1]==a[2])&&(a[2]==a[3])){cout << a[0] << a[1] << a[2] << a[3] << " - " << a[0] << a[1] << a[2] << a[3] << " = 0000" << endl;return 0;}else{while(1){sort(a,a+4,cmp);cout << a[0] << a[1] << a[2] << a[3] << " - " ;for(int i=0;i<4;i++){b[i] = a[i];}sort(b,b+4);cout << b[0] << b[1] << b[2] << b[3] << " = ";for(int i=3;i>=0;i--){if(a[i]-b[i]<0){c[i] = a[i] + 10 - b[i];a[i-1]--;}else{c[i] = a[i] - b[i];}}  cout << c[0] << c[1] << c[2] << c[3] << endl;if(c[0]==6 && c[1]==1 && c[2]==7 && c[3]==4){    break;}else{for(int i=0;i<4;i++){a[i] = c[i];}}}}return 0;
}

1070 Mooncake (25 分)

算法标签: 模拟 + 排序

#include<bits/stdc++.h>
using namespace std;
int N,D;
int maxn = 1005;
typedef struct mooncake{double num;double price;double singleprice;
}mooncake;
bool cmp(mooncake m1,mooncake m2){return m1.singleprice > m2.singleprice;
}
int main(){cin >> N >> D;mooncake m[N];for(int i=0;i<N;i++){cin >> m[i].num;}for(int i=0;i<N;i++){cin >> m[i].price;}for(int i=0;i<N;i++){m[i].singleprice = (m[i].price * 1.0) /(m[i].num * 1.0);}sort(m,m+N,cmp);int site = 0;double totalprice = 0.0;while(D!=0){if(m[site].num <= D){D -= m[site].num;totalprice = totalprice + m[site].price;}else{totalprice = totalprice + m[site].singleprice * D;D = 0;}site++;}printf("%.2lf\n",totalprice);return 0;
}

1071 Speech Patterns (25 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1048576 + 5;
char s[maxn];
map<string,int> mp;
int iscase(char ch){if((ch>='A' && ch<='Z') || (ch>='a' && ch<='z')){return 1;}else{return 0;}
}
int check(char ch){if(iscase(ch) || (ch>='0' && ch<='9')){return 1;}else{return 0;}
}int main(){string tmp;getline(cin,tmp);strcpy(s,tmp.c_str());int len = strlen(s);string t = "";for(int i=0;i<len;i++){if(check(s[i])){if(iscase(s[i])){t += tolower(s[i]); }else{t += s[i];}}if(!check(s[i]) || i==len-1){if(t.length()!=0){mp[t]++;}t.clear();}}map<string,int>::iterator it = mp.begin();string ans;int res = -1;for(;it!=mp.end();it++){if(it->second>res){res = it->second;ans = it->first;}}cout << ans << " " << res << endl;return 0;
}

1072 Gas Station (30 分)

算法标签: 图论 + Dijkstra
注意点: 参考P1003/P1018/P030，解法相同

#include<bits/stdc++.h>
using namespace std;
int N,M,K,D;double ans = -1;
double average = 1e8;
int res = -1;struct node{int v;int dis;
};vector<node> G[1050];
int dp[1050];
bool vis[1050];void dijstra(int s){memset(vis,0,sizeof(vis));fill(dp+1,dp+N+M+1,INT_MAX);dp[s] = 0;for(int i=1;i<=N+M;i++){int u = -1;int minnum = INT_MAX;for(int j=1;j<=N+M;j++){if(!vis[j] && dp[j]<minnum){minnum = dp[j];u = j;}}if(u==-1){return;}vis[u] = true;for(int j=0;j<G[u].size();j++){int v = G[u][j].v;int dis = G[u][j].dis;if(!vis[v]){if(dp[u]+dis<dp[v]){dp[v] = dp[u]+dis; }}}}
}void deal(int M){bool f = true;int mindis = INT_MAX;for(int i=1;i<=N;i++){if(dp[i]<mindis){mindis = dp[i];}if(dp[i]>D){f = false;break;}}if(f){double tmp = 0.0;for(int i=1;i<=N;i++){tmp += 1.0*dp[i]/N;}if(mindis>ans){ans = mindis;average = tmp;res = M;}else if(mindis==ans){if(tmp<average){average = tmp;res = M;   }   }}
}int main(){scanf("%d%d%d%d",&N,&M,&K,&D);for(int i=0;i<K;i++){int p1 = 0;int p2 = 0;int dist;string t1;string t2;cin >> t1 >> t2;scanf("%d",&dist);if(t1[0] == 'G'){for(int j=1;j<t1.size();j++){p1 = p1*10 + t1[j]-'0';}p1 = p1 + N;}else{for(int j=0;j<t1.size();j++){p1 = p1*10 + t1[j]-'0';}}if(t2[0] == 'G'){for(int j=1;j<t2.size();j++){p2 = p2*10 + t2[j]-'0';}p2 = p2 + N;}else{for(int j=0;j<t2.size();j++){p2 = p2*10 + t2[j]-'0';}}node temp;temp.dis = dist;temp.v = p2;G[p1].push_back(temp);temp.v = p1;G[p2].push_back(temp);}for(int i=N+1;i<=N+M;i++){dijstra(i);deal(i);}if(res == -1){printf("No Solution\n");}else{printf("G%d\n",res-N);printf("%.1lf %.1lf\n",ans,average);}return 0;
}

1073 Scientific Notation (20 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;int main(){string str;cin >> str;int site = str.find('E');string n = str.substr(1,site-1);string e = str.substr(site+1);int en = stoi(e);if(str[0] == '-'){cout << "-";}if(en>=0){cout << str[1];int j=2,cnt=0;for(j=2;j<site && cnt<en;j++){if(str[j]=='.'){continue;}else{cout << str[j];cnt++;}} if(j==site){for(int i=0;i<en-cnt;i++){cout << "0" ; }}else{cout <<".";for(int i=j;j<site;j++){cout << str[j];}}}else{cout << "0.";for(int i=0;i<abs(en)-1;i++){cout << "0";}cout << str[1];for(int j=2;j<site;j++){if(str[j]=='.'){continue;}else{cout << str[j];}}}return 0;
}

1074 Reversing Linked List (25 分)

算法标签: 模拟 + 链表

#include<bits/stdc++.h>
using namespace std;
int start,N,K;
const int maxn = 1e5+5;
int e_data[maxn];
int eadd[maxn];
int listnum[maxn];int main(){cin.tie(0);cout.tie(0);cin >> start >> N >> K;for(int i=0;i<N;i++){int addr;cin >> addr;cin >> e_data[addr] >> eadd[addr];}int nextadd = start;int count = 0;while(nextadd!=-1){listnum[count++] = nextadd;nextadd = eadd[nextadd];}for(int i=0;i<(count-count%K);i+=K){reverse(listnum+i,listnum+i+K);}for(int i=0;i<count;i++){if(i!=count-1){printf("%05d %d %05d\n",listnum[i],e_data[listnum[i]],listnum[i+1]);}else{printf("%05d %d %d\n",listnum[i],e_data[listnum[i]],-1);}}return 0;
}

1075 PAT Judge (25 分)

算法标签: 模拟 + 排序

#include<bits/stdc++.h>
using namespace std;
int score[10];const int maxn = 1e5+5;struct student{int total_score=0;int score_get[6] = {-1,-1,-1,-1,-1,-1};int rank;int perfect = 0;int id;
};student stu[maxn];bool cmp(student s1,student s2){if(s1.score_get[0]!=s2.score_get[0]){return s1.score_get[0] > s2.score_get[0];}else if(s1.total_score!=s2.total_score){return s1.total_score > s2.total_score;}else if(s1.perfect!=s2.perfect){return s1.perfect > s2.perfect;}else{return s1.id < s2.id;}
}int main(){int N,M,K;scanf("%d%d%d",&N,&K,&M);for(int i=1;i<=K;i++){scanf("%d",&score[i]);}for(int i=0;i<M;i++){int id,pro_id,score_num;scanf("%d%d%d",&id,&pro_id,&score_num);stu[id].id = id;if(score_num==-1){if(stu[id].score_get[pro_id] == -1){stu[id].score_get[pro_id] = -2;}}else{stu[id].score_get[0] = 1;if(stu[id].score_get[pro_id]<score_num){stu[id].score_get[pro_id] = max(stu[id].score_get[pro_id],score_num);}int t = 0;int count = 0;for(int j=1;j<=K;j++){if(stu[id].score_get[j]!=-1 && stu[id].score_get[j]!=-2){count += stu[id].score_get[j];}if(stu[id].score_get[j] == score[j]){t++;}}stu[id].perfect = t;stu[id].total_score = count;}}sort(stu,stu+maxn,cmp);stu[0].rank = 1;for(int i=0;i<maxn;i++){if(stu[i].score_get[0] == -1){break;}else{if(stu[i].total_score == stu[i+1].total_score){stu[i+1].rank = stu[i].rank;}else{stu[i+1].rank = i+2;}printf("%d %05d %d",stu[i].rank,stu[i].id,stu[i].total_score);for(int j=1;j<=K;j++){if(stu[i].score_get[j]==-1){printf(" -");}else if(stu[i].score_get[j] == -2){printf(" 0");}else{printf(" %d",stu[i].score_get[j]);}}printf("\n");}}return 0;
}

1076 Forwards on Weibo (30 分)

算法标签: 图论 + BFS
注意点: 从查询的源点出发，BFS遍历不超过L层，统计L层的信息即可

#include<bits/stdc++.h>
using namespace std;
int a[1005][1005];
bool f[1005];
struct node{int layer;int number;
};
int N,L;int BFS(int u){node s;f[u] = true;s.layer = 0;s.number = u;queue<node> q;q.push(s);int total = -1;while(!q.empty()){node front = q.front();q.pop();if(front.layer>L){break;}else{total++;}for(int i=1;i<=N;i++){if(!f[i] && a[front.number][i]){f[i] = true;node tmp;tmp.layer = front.layer+1;tmp.number = i;q.push(tmp);}}}return total;
}
int main(){cin >> N >> L;for(int i=1;i<=N;i++){int k;scanf("%d",&k);for(int j=0;j<k;j++){int v;scanf("%d",&v);a[v][i] = 1;}}int k;scanf("%d",&k);for(int i=0;i<k;i++){int u;scanf("%d",&u);memset(f,0,sizeof(f));printf("%d\n",BFS(u));}return 0;
}

1077 Kuchiguse (20 分)

算法标签: 模拟 + 字符串
注意点: 从每个字符串的末尾开始逆序枚举，求解最长的公共后缀

#include<bits/stdc++.h>
using namespace std;
string s[105];
int main(){int N;cin >> N;getchar();int minn = INT_MAX;for(int i=0;i<N;i++){getline(cin,s[i]);int len = s[i].size();reverse(s[i].begin(),s[i].end());minn = min(minn,len);}int cnt = 0;for(int i=0;i<minn;i++){bool f = true;char c = s[0][i];for(int j=0;j<N;j++){if(c == s[j][i]){continue;}else{f = false;break;}}if(f){cnt++;}else{break;}}if(cnt){for(int i=cnt-1;i>=0;i--){cout << s[0][i];} cout << endl;}else{cout << "nai" << endl; }return 0;
}

1078 Hashing (25 分)

算法标签: 模拟 + 哈希
注意点: 模拟哈希表的工作过程，哈希函数本题采用的是除留取余法，解决冲突采用的是平方试探法

#include<bits/stdc++.h>
using namespace std;const int maxn = 1e5+5;
bool p[maxn]={1,1,0};
int a[maxn];
bool hashT[maxn];void init(){for(int i=2;i<maxn;i++){if(!p[i]){for(int j=2*i;j<maxn;j+=i){p[j] = 1;}}}
}int main(){init();int msize;int N;int Tsize;scanf("%d%d",&msize,&N);for(int i=0;i<N;i++){scanf("%d",&a[i]);}if(!p[msize]){Tsize = msize;}else{for(int i=msize+1;;i++){if(!p[i]){Tsize = i;break;}}}for(int i=0;i<N;i++){int key = a[i] % Tsize;if(!hashT[key]){hashT[key] = true;if(i==0){printf("%d",key);}else{printf(" %d",key);}}else{bool f = false;for(int step=1;step<Tsize;step++){int k = (a[i]+step*step)%Tsize;if(!hashT[k]){hashT[k] = true;if(i==0){printf("%d",k);}else{printf(" %d",k);}f = true;break;}} if(!f){if(i==0){printf("-");}else{printf(" -");}}}}return 0;
}

1079 Total Sales of Supply Chain (25 分)

算法标签: 树，二叉树 + DFS
注意点: 参考P1021/P1053，解法相同

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
vector<int> child[maxn];
int a[maxn];
int N;
double P;
double r;
bool hashT[maxn];
int mindepth = INT_MAX;
int minnum = 0;
double res = 0.0;void dfs(int depth,int root){if(child[root].size()==0){res += P*pow(1.0+r/100.0,depth)*a[root];return;}for(int i=0;i<child[root].size();i++){dfs(depth+1,child[root][i]);}
}int main(){cin >> N >> P >> r;for(int i=0;i<N;i++){int k;scanf("%d",&k);if(k==0){scanf("%d",&a[i]);}else{for(int j=0;j<k;j++){int t;scanf("%d",&t);child[i].push_back(t);hashT[t] = 1;}}}int root;for(int i=0;i<N;i++){if(!hashT[i]){root = i;break;}}dfs(0,root);printf("%.1lf\n",res);return 0;
}

1080 Graduate Admission (30 分)

算法标签: 模拟 + 排序

#include<bits/stdc++.h>
using namespace std;
int cruit_num[105];
struct student{int GE;int GI;int final;int rank;int choice[5]; int id;
};
student stu[40005];
int choice[105][40005];
int cnt[105];bool cmp(student s1,student s2){if(s1.final!=s2.final){return s1.final > s2.final;}else{return s1.GE > s2.GE;}
}int main(){int N,M,K;scanf("%d%d%d",&N,&M,&K);for(int i=0;i<M;i++){scanf("%d",&cruit_num[i]);}for(int i=0;i<N;i++){scanf("%d%d",&stu[i].GE,&stu[i].GI);stu[i].final = stu[i].GE + stu[i].GI;for(int j=0;j<K;j++){scanf("%d",&stu[i].choice[j]);}stu[i].id = i;}sort(stu,stu+N,cmp);stu[0].rank = 1;for(int i=1;i<N;i++){if(stu[i].final == stu[i-1].final && stu[i].GE == stu[i-1].GE){stu[i].rank = stu[i-1].rank;}else{stu[i].rank = i+1;}}for(int i=0;i<N;i++){for(int j=0;j<K;j++){int id = stu[i].choice[j]; if(cruit_num[id]>0){cruit_num[id]--;choice[id][cnt[id]++] = i;break;}else if(cruit_num[id]==0 && cnt[id]>0 && stu[choice[id][cnt[id]-1]].rank == stu[i].rank){choice[id][cnt[id]++] = i;break;}}}for(int i=0;i<M;i++){if(cnt[i]>0){for(int j=0;j<cnt[i];j++){choice[i][j] = stu[choice[i][j]].id;}}}for(int i=0;i<M;i++){sort(choice[i],choice[i]+cnt[i]); }for(int i=0;i<M;i++){if(cnt[i]>0){for(int j=0;j<cnt[i];j++){if(j==0){printf("%d",choice[i][j]);}else{printf(" %d",choice[i][j]);}}  }printf("\n");}return 0;
}

1081 Rational Sum (20 分)

算法标签: 模拟 + 分数类

#include<bits/stdc++.h>
using namespace std;struct fraction{long long up;long long down;
};
fraction f[105];long long gcd(long long a,long long b){if(a<b){swap(a,b);}if(b==0){return a;}else{return gcd(b,a%b);}
}fraction Reduction(fraction f1){if(f1.down<0){f1.up = -f1.up;f1.down = -f1.down;}if(f1.up == 0){f1.down = 1;}else{long long g = gcd(abs(f1.up),abs(f1.down));f1.down /= g;f1.up /= g;   }return f1;
}fraction Add(fraction f1,fraction f2){fraction res;res.down = f1.down*f2.down;res.up = f1.down*f2.up + f2.down*f1.up;res = Reduction(res);return res;
}int main(){int n;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lld/%lld",&f[i].up,&f[i].down);f[i] = Reduction(f[i]);}fraction sum;sum.up = 0;sum.down = 1;for(int i=0;i<n;i++){sum = Add(sum,f[i]);}if(sum.down == 1){printf("%lld\n",sum.up);}else if(sum.up>sum.down){printf("%lld %lld/%lld\n",sum.up/sum.down,sum.up%sum.down,sum.down);}else{printf("%lld/%lld\n",sum.up,sum.down);}return 0;
}

1082 Read Number in Chinese (25 分)

算法标签: 模拟 + 字符串
注意点: 一道好题！

#include<bits/stdc++.h>
using namespace std;
string out[] = {"","Shi","Bai","Qian","Wan","Yi"};
string digit[] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
int main(){int n;cin >> n;if(n<0){cout << "Fu ";}n = abs(n);if(n==0){cout << "ling" << endl;return 0;}char str[15];sprintf(str,"%d",n);int len = strlen(str);int left = 0;int right = len-1;while(left+4<=right){right -= 4;}while(left<len){bool f = false;while(left<=right){if(str[left] == '0'){f = true;}else{if(f){cout << " ling";f = false;}if(left>0){cout << " " << digit[str[left]-'0'];}else{cout << digit[str[left]-'0'];}if(right!=left){cout << " " << out[right-left];}}left++;}int k = (len-left)/4 + 3;if(left!=len){cout << " " << out[k];}right += 4;}return 0;
}

1083 List Grades (25 分)

算法标签: 模拟 + 排序

#include<bits/stdc++.h>
using namespace std;
struct student{char name[15];char course[15];int score;
};
const int maxn = 1e4+5;
student s[maxn];
bool cmp(student s1,student s2){return s1.score > s2.score;
}
int main(){int N;scanf("%d",&N);for(int i=0;i<N;i++){scanf("%s%s%d",s[i].name,s[i].course,&s[i].score);}int s1,e1;scanf("%d%d",&s1,&e1);sort(s,s+N,cmp);if(s1>e1){swap(s1,e1);}int cnt = 0;for(int i=0;i<N;i++){if(s[i].score>=s1 && s[i].score<=e1){cnt++;printf("%s %s\n",s[i].name,s[i].course);}}if(!cnt){printf("NONE\n");}return 0;
}

1084 Broken Keyboard (20 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;
int a[55];
int isUpper(char c){if(c>='A' && c<='Z'){return 1;}else{return 0;}
}int isdigit(char c){if(c>='0' && c<='9'){return 1;}else{return 0;}
}int isLower(char c){if(c>='a' && c<='z'){return 1;}else{return 0;}
}int main(){string s1,s2;cin >> s1 >> s2;int j=0;for(int i=0;i<s1.size();i++){if(s1[i]!=s2[j]){if(s1[i]=='_'){if(!a[10]){a[10] = 1;cout << s1[i];    }}else if(isdigit(s1[i])){if(!a[s1[i]-'0']){a[s1[i]-'0'] = 1;cout << s1[i];}}else if(isLower(s1[i])){if(!a[11+s1[i]-'a']){a[11+s1[i]-'a'] = 1;char alpha = s1[i] - 'a' + 'A';cout << alpha;}}else if(isUpper(s1[i])){if(!a[11+s1[i]-'A']){a[11+s1[i]-'A'] = 1;cout << s1[i]; } }}else{j++;}}return 0;
}

1085 Perfect Sequence (25 分)

算法标签: 双指针
注意点: 令指针i指向序列头，指针j指向a[i]*p,统计j到i的最大值，由于指针的顺序性，因此总体复杂度为 O ( n l o g n ) O(nlogn) O(nlogn)

#include "iostream"
#include "string"#include "algorithm"using namespace std;int main(){int N,maxCount=1;double p,a[100010];scanf("%d %lf",&N,&p);for (int i=0;i<N;i++){scanf("%lf",&a[i]);}sort(a,a+N);for(int i=0;i<N;i++){double beginNum=a[i];for(int j=i+maxCount;j<N;j++){double endNum=a[j];if(beginNum*p<endNum){break;}if (j-i+1>maxCount){maxCount=j-i+1;}}}cout<<maxCount;
}

1086 Tree Traversals Again (25 分)

算法标签: 树，二叉树 + DFS
注意点: 参考P1021/P1053/P1079，解法相同

#include<bits/stdc++.h>
using namespace std;
int pre[105];
int in[105];
int post[105];stack<int> s;
struct node{int data;node* lchild;node* rchild;
};
node* root = NULL;
node* newNode(int v){node* Node = new node;Node->data = v;Node->lchild = NULL;Node->rchild = NULL; return Node;
}
node* Create(int preL,int preR,int inL,int inR){if(preL>preR){return NULL;}node* Node = newNode(pre[preL]);int k;for(int i=inL;i<=inR;i++){if(in[i]==Node->data){k = i;break;}}int numleft = k-inL;Node->lchild = Create(preL+1,preL+numleft,inL,k-1);Node->rchild = Create(preL+numleft+1,preR,k+1,inR);return Node;
}
int cnt3 = 0;
void postorder(node* root){if(root==NULL){return;}postorder(root->lchild);postorder(root->rchild);if(cnt3==0){printf("%d",root->data);}else{printf(" %d",root->data);}cnt3++;
}
int main(){int N;int cnt1 = 0;int cnt2 = 0;cin >> N;for(int i=0;i<2*N;i++){string s1;cin >> s1;if(s1=="Push"){int t;cin >> t;pre[cnt1++] =  t;s.push(t);}else{in[cnt2++] = s.top();s.pop();}}root = Create(0,N-1,0,N-1);postorder(root);return 0;
}

1087 All Roads Lead to Rome (30 分)

算法标签: 图论 + Dijkstra + DFS
注意点: 参考P1003/P1018/P1030/P1072，解法相同

#include<bits/stdc++.h>
using namespace std;int N,K;
string city1;
map<string,int> strtonum;
map<int,string> numtostr;int w[205];
int weight[205];
int num[205];
int D;
bool vis[205];
int dp[205];
int cnt[205];
int pre[205];struct node{int dis;int v;
};
vector<node> G[205];void dfs(int v){if(v == 0){cout << numtostr[v];return;} dfs(pre[v]);printf("->");cout << numtostr[v];
}
void dijstra(int s){memset(vis,0,sizeof(vis));fill(dp,dp+N,INT_MAX);fill(num,num+N,1);dp[s] = 0;for(int i=0;i<N;i++){int u = -1;int minnum = INT_MAX;for(int j=0;j<N;j++){if(!vis[j] && dp[j]<minnum){minnum = dp[j];u = j;}}vis[u] = true;for(int j=0;j<G[u].size();j++){int v = G[u][j].v;int dis = G[u][j].dis;if(!vis[v]){if(dp[u]+dis<dp[v]){dp[v] = dp[u]+dis;num[v] = num[u];weight[v] = weight[u] + w[v];cnt[v] = cnt[u] + 1;pre[v] = u;}else if(dp[u]+dis==dp[v]){num[v] += num[u];if(weight[v]<weight[u]+w[v]){weight[v] = weight[u] + w[v];cnt[v] = cnt[u] + 1;pre[v] = u;}else if(weight[v]==weight[u]+w[v]){if(cnt[v]>cnt[u]+1){cnt[v] = cnt[u]+1;pre[v] = u;}}}}}}
}
int main(){cin >> N >> K >> city1;strtonum[city1] = 0;numtostr[0] = city1;for(int i=1;i<=N-1;i++){string t;cin >> t;cin >> w[i];if(t=="ROM"){D = i;}strtonum[t] = i;numtostr[i] = t;}for(int i=0;i<K;i++){string c1,c2;int dis;cin >> c1 >> c2 >> dis;node temp;int n1 = strtonum[c1];int n2 = strtonum[c2];temp.v = n2;temp.dis = dis;G[n1].push_back(temp);temp.v = n1;G[n2].push_back(temp);}dijstra(0);printf("%d %d %d %d\n",num[D],dp[D],weight[D],weight[D]/cnt[D]);dfs(D);return 0;
}

1088 Rational Arithmetic (20 分)

算法标签: 模拟 + 数论

#include<bits/stdc++.h>
using namespace std;
long long gcd(long long a,long long b){long long tmp;if(a<b){tmp = a;a = b;b = tmp;}return a%b? gcd(b,a%b) : b;
}
long long lcd(long long a,long long b){long long g = gcd(a,b);return(1LL*(a/g)*(b/g));
}void process(long long a,long long b,int T){bool flag = false,flag1 = false,flag2=false;if(a<0){flag1 = true;}if(b<0){flag2 = true;}if(flag1 && flag2){a = -a;b = -b;flag = false;}else if(flag1 && !flag2){a = -a;flag = true;}else if(!flag1 && flag2){b = -b;flag = true;}if(a==0 && b!=0){cout << "0";}else if(b==0 && T==1){cout << "Inf";}else{long long g = gcd(a,b);a = a/g;b = b/g;if(flag){cout << "(-";}if(a>=b){int p = a/b;int q = a%b;if(q!=0){cout << p <<" " << q << "/" << b;}else{cout << p ;}}else{cout << a << "/" << b;}if(flag){cout <<")";}}
}int main(){char str1[50],str2[50];cin >> str1 >> str2;long long a,b,c,d;sscanf(str1,"%lld/%lld",&b,&a);sscanf(str2,"%lld/%lld",&d,&c);long long plus1 = b*c + a*d;long long plus2 = a*c;long long minus1 = b*c - a*d;long long minus2 = a*c;long long mutl1 = b*d;long long mutl2 = a*c;long long div1 = b*c;long long div2 = a*d;process(b,a,0);cout << " + " ;process(d,c,0);cout << " = " ;process(plus1,plus2,0);cout << endl;process(b,a,0);cout << " - " ;process(d,c,0);cout << " = " ;process(minus1,minus2,0);cout << endl;process(b,a,0);cout << " * " ;process(d,c,0);cout << " = " ;process(mutl1,mutl2,0);cout << endl;process(b,a,0);cout << " / " ;process(d,c,0);cout << " = " ;process(div1,div2,1);cout << endl;return 0;
}

1089 Insert or Merge (25 分)

算法标签: 模拟 + 排序
注意点: 要判断是插入排序还是归并排序（或是P1098中的堆排序），我们需要对每个排序过程十分清晰，插入排序是指有一个序列，每次从前往后将一个无序的元素插入到有序的数列中，因此，插入排序必定是前半部分有序，后半部分无序且和输入的序列一致；接下来看看归并排序，归并排序是两两一组合并，因此我们假设数据元素的位置为i，那么第n轮排序，元素必定位于 i 2 n + [ 2 n , 2 n + 1 ] \frac{i}{2^n}+[2^n,2^{n+1}] 2ni+[2n,2n+1]之间；最后看堆排序，先建堆，然后每一轮堆排序都是上滤的过程，不断交换子结点与父结点

#include<bits/stdc++.h>
using namespace std;
int N;
const int maxn = 1e2+5;
int a[maxn];
int b[maxn];int main(){cin >> N;for(int i=0;i<N;i++){cin >> a[i];}for(int i=0;i<N;i++){cin >> b[i];}int j=0;while(j<N-1 && b[j]<=b[j+1]){j++;}int site1 = j;j++;while(j<N && a[j]==b[j]){j++;}if(j==N){cout << "Insertion Sort" << endl;sort(b,b+site1+2);for(int i=0;i<N;i++){if(i!=N-1){cout << b[i] << " ";}else{cout << b[i];}}}else{cout <<"Merge Sort" << endl;int k = 1;while(1){k *= 2;for(int i=0;i<N-N%k;i+=k){sort(a+i,a+i+k);} sort(a+N-N%k,a+N);bool flag = true;for(int i=0;i<N;i++){if(a[i]!=b[i]){flag = false;break;}}if(flag){break;}}k *= 2;for(int i=0;i<N-N%k;i+=k){sort(a+i,a+i+k);}    sort(a+N-N%k,a+N);for(int i=0;i<N;i++){if(i!=N-1){cout << a[i] << " ";}else{cout << a[i];}}}return 0;
}

1090 Highest Price in Supply Chain (25 分)

算法标签: 树，二叉树 + DFS
注意点: 参考P1021/P1053/P1079/P1086，解法相同

#include<bits/stdc++.h>
using namespace std;
int N;
double P;
double r;
const int maxn = 1e5+5;
int a[maxn];
struct node{int data;vector<int> child;
};
node n[maxn];int index_num = 0;
int newnode(int v){n[index_num].data = v;return index_num++;
}
int maxnum = 0;
int maxdepth = -1;void dfs(int depth,int now){if(n[now].child.size()==0){if(depth>maxdepth){maxdepth=depth;maxnum = 1;}else if(depth==maxdepth){maxnum++;}return;}for(int i=0;i<n[now].child.size();i++){dfs(depth+1,n[now].child[i]);}
}int main(){cin >> N >> P >> r;for(int i=0;i<N;i++){scanf("%d",&a[i]);}int root;for(int i=0;i<N;i++){int now = newnode(i);if(a[now]!=-1){n[a[now]].child.push_back(now);}else{root = now;}}dfs(0,root);double res = P*pow(1+r/100.0,maxdepth);printf("%.2lf %d\n",res,maxnum);return 0;
}

1091 Acute Stroke (30 分)

算法标签: BFS
注意点: 这道题是一道BFS模板题，由于三维方块有6个方向，因此BFS可遍历的宽度为6，每次BFS时需判断是否出界、是否访问过，BFS是队列结构，初始队列内为开始访问的起点，整个队列全部为空时结束BFS的访问

#include<bits/stdc++.h>
using namespace std;
bool f[1350][135][65];
int a[1350][135][65];
int M,N,L,T;
int xx[]={0,0,1,0,0,-1};
int yy[]={0,1,0,0,-1,0};
int zz[]={1,0,0,-1,0,0};
struct node{int x,y,z;
};
node temp;
queue<node> q;int check(int x,int y,int z){if(x>=1 && x<=M && y>=1 && y<=N && z>=1 && z<=L){return 1;}else{return 0;}
}int BFS(){int cnt = 0;while(!q.empty()){node front_node = q.front();q.pop();cnt++;for(int i=0;i<6;i++){temp.x = front_node.x + xx[i];temp.y = front_node.y + yy[i];temp.z = front_node.z + zz[i];if(check(temp.x,temp.y,temp.z) && !f[temp.x][temp.y][temp.z] && a[temp.x][temp.y][temp.z]==1){f[temp.x][temp.y][temp.z] = 1;q.push(temp);}}} if(cnt>=T){return cnt;    }else{return 0;}
}
int main(){cin >> M >> N >> L >> T;for(int z=1;z<=L;z++){for(int x=1;x<=M;x++){for(int y=1;y<=N;y++){scanf("%d",&a[x][y][z]);}}}int ans = 0;for(int z=1;z<=L;z++){for(int x=1;x<=M;x++){for(int y=1;y<=N;y++){if(a[x][y][z] == 1 && !f[x][y][z]){temp.x = x;temp.y = y;temp.z = z;f[x][y][z] = 1;q.push(temp);ans += BFS();}}}}cout << ans << endl;return 0;
}

1092 To Buy or Not to Buy (20 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;
string str1,str2;
int a[105];
int b[105];
int isupper(char c){if(c>='A' && c<='Z'){return 1;}else{return 0;}
}
int islower(char c){if(c>='a' && c<='z'){return 1;}else{return 0;}
}
int isdigit(char c){if(c>='0' && c<='9'){return 1;}else{return 0;}
}
int main(){cin >> str1 >> str2;for(int i=0;i<str1.size();i++){if(isdigit(str1[i])){a[str1[i]-'0']++;}else if(isupper(str1[i])){a[str1[i]-'A'+10]++;}else if(islower(str1[i])){a[str1[i]-'a'+40]++;}}for(int i=0;i<str2.size();i++){if(isdigit(str2[i])){b[str2[i]-'0']++;}else if(isupper(str2[i])){b[str2[i]-'A'+10]++;}else if(islower(str2[i])){b[str2[i]-'a'+40]++;}}bool flag = true;int rest = 0;int count1=0,count2=0;for(int i=0;i<100;i++){count1 += a[i];count2 += b[i];if(a[i]<b[i]){flag = false;rest += (b[i]-a[i]);}}if(flag==false){cout << "No ";cout << rest << endl;}else{cout << "Yes ";cout << count1-count2 << endl;}return 0;
}

1093 Count PAT’s (25 分)

算法标签: DP
注意点: 把字符串分成3个部分,left/mid/right，若mid=‘A’,那么这个点处PAT的数量等于 l e f t ( P ) ∗ r i g h t ( T ) left(P)*right(T) left(P)∗right(T)，因此 ∑ \sum ∑累加即可

#include<bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
long long result;
const int maxn = 1e5+5;
string str;int main(){cin >> str;int Tcount = 0;int Pcount = 0;for(int i=0;i<str.size();i++){if(str[i]=='T'){Tcount++;}}for(int i=0;i<str.size();i++){if(str[i]=='T'){Tcount--;}else if(str[i]=='P'){Pcount++;}else if(str[i]=='A'){result = (result + Tcount*Pcount)%M;}}cout << result << endl;return 0;
}

1094 The Largest Generation (25 分)

算法标签: 树，二叉树 + DFS
注意点: 建树，做DFS遍历，统计每一层上的结点数，输出最大值及其层号

#include<bits/stdc++.h>
using namespace std;int N,M;
int a[105];
vector<int> child[105];
int maxdepth = -1;void dfs(int depth,int root){a[depth]++;if(depth>maxdepth){maxdepth = depth;}  for(int i=0;i<child[root].size();i++){dfs(depth+1,child[root][i]);}
}int main(){memset(a,0,sizeof(a));cin >> N >> M;for(int i=0;i<M;i++){int id;cin >> id;int k;cin >> k;for(int j=0;j<k;j++){int t;cin >> t;child[id].push_back(t);}}    int root=1;dfs(1,root);int site;int maxnum =-1;for(int i=1;i<=maxdepth;i++){if(a[i]>maxnum){maxnum = a[i];site = i;}}printf("%d %d\n",maxnum,site);return 0;
}

1095 Cars on Campus (30 分)

算法标签: 大模拟 + 排序

#include<bits/stdc++.h>
using namespace std;
struct record{char id[9];int time;bool status;
};
const int maxn = 1e4+5;
record r[maxn];
record valid[maxn];
int a[maxn];
map<string,int> mp;bool cmp(record r1,record r2){if(strcmp(r1.id,r2.id)!=0){return strcmp(r1.id,r2.id)<0;}else{return r1.time < r2.time;}
}bool cmp1(record r1,record r2){return r1.time < r2.time;
}int main(){int N,K;scanf("%d%d",&N,&K);for(int i=0;i<N;i++){int h,m,s;char status[10];scanf("%s%d:%d:%d%s",r[i].id,&h,&m,&s,status);r[i].time = h*3600 + m*60 + s;if(strcmp(status,"in")==0){r[i].status = 1;}else{r[i].status = 0;}} sort(r,r+N,cmp);int cnt = 0;for(int i=0;i<N;i++){if(strcmp(r[i].id,r[i+1].id)==0){if(r[i].status == 1 && r[i+1].status == 0){strcpy(valid[cnt].id,r[i].id);valid[cnt].time = r[i].time;valid[cnt].status = r[i].status;cnt++;strcpy(valid[cnt].id,r[i+1].id);valid[cnt].time = r[i+1].time;valid[cnt].status = r[i+1].status;cnt++;mp[r[i].id] += (r[i+1].time-r[i].time);}}}sort(valid,valid+cnt,cmp1);int index = 0;int count = 0;for(int i=0;i<K;i++){int h,m,s;scanf("%d:%d:%d",&h,&m,&s);int time = h*3600 + m*60 + s;while(valid[index].time<=time && index<cnt){if(valid[index].status == 1){count++;}else{count--;}index++;}printf("%d\n",count);}int maxmum = -1;for(auto it=mp.begin();it!=mp.end();it++){if(it->second>maxmum){maxmum = it->second;}}int res;for(auto it=mp.begin();it!=mp.end();it++){if(it->second == maxmum){printf("%s ",it->first.c_str());res = it->second;}}printf("%02d:%02d:%02d\n",res/3600,res/60%60,res%60);return 0;
}

1096 Consecutive Factors (20 分)

算法标签: 数论

#include<bits/stdc++.h>
using namespace std;int main(){long long n;scanf("%lld",&n);int maxlen = -1;int site;for(int i=2;i<=sqrt(n);i++){long long tmp = n;int j=i;while(1){if(tmp%j==0){tmp/=j;j++;}else{break;}}if(j-i>maxlen && j!=i){maxlen = j-i;site = i;}}if(maxlen==-1){printf("1\n");printf("%lld\n",n);}else{printf("%d\n",maxlen);for(int i=site;i<site+maxlen;i++){if(i==site){printf("%d",i);}else{printf("*%d",i);}}printf("\n");}return 0;
}

1097 Deduplication on a Linked List (25 分)

算法标签: 模拟 + 链表

#include<bits/stdc++.h>
using namespace std;
struct node{int value;int next;
};
const int maxn = 1e5+5;
node n[maxn];int ori_add[maxn];
int ori_value[maxn];
int post_add[maxn];
int post_value[maxn];
int cnt1 = 0;
int cnt2 = 0;
set<int> s;int main(){int start;int N;cin >> start >> N;for(int i=0;i<N;i++){int y;scanf("%d",&y);scanf("%d%d",&n[y].value,&n[y].next);}while(start!=-1){if(s.find(abs(n[start].value))!=s.end()){post_add[cnt2] = start;post_value[cnt2] = n[start].value;cnt2++; }else{ori_add[cnt1] = start;ori_value[cnt1] = n[start].value;s.insert(abs(n[start].value));cnt1++;}start = n[start].next;} for(int i=0;i<cnt1;i++){if(i!=cnt1-1){printf("%05d %d %05d\n",ori_add[i],ori_value[i],ori_add[i+1]);}else{printf("%05d %d -1\n",ori_add[i],ori_value[i]);}}for(int i=0;i<cnt2;i++){if(i!=cnt2-1){printf("%05d %d %05d\n",post_add[i],post_value[i],post_add[i+1]);}else{printf("%05d %d -1\n",post_add[i],post_value[i],post_add[i+1]);}}return 0;
}

1098 Insertion or Heap Sort (25 分)

算法标签: 模拟 + 排序
注意点: 可参考P1089的解法

#include<bits/stdc++.h>
using namespace std;
int res[105];
int a[105];
int b[105];
int c[105];
int N;
int equal(int a[],int b[]){int f = 1;for(int i=0;i<N;i++){if(a[i]==b[i]){continue;}else{f = 0;break;}}return f;
}void insertionsort(int a[],int res[]){for(int i=1;i<=N;i++){sort(a+1,a+i+1);int f = equal(a,res);if(i!=1 && f){sort(a+1,a+i+2);printf("Insertion Sort\n");for(int j=1;j<=N;j++){if(j==1){printf("%d",a[j]);}else{printf(" %d",a[j]); }}printf("\n");break;}}
}void down(int left,int right){int i = left;int j = 2*i;while(j<=right){if(j+1<=right && c[j+1]>c[j]){j = j+1;}if(c[i]<c[j]){swap(c[j],c[i]);i = j;j = 2*i;}else{break;}}
}void heapsort(int a[],int res[]){for(int i=N/2;i>=1;i--){down(i,N);}for(int i=N;i>=1;i--){swap(a[1],a[i]);down(1,i-1);int f = equal(a,res);if(f){printf("Heap Sort\n");swap(a[1],a[i-1]);down(1,i-2);for(int j=1;j<=N;j++){if(j==1){printf("%d",a[j]); }else{printf(" %d",a[j]);}}printf("\n");break;}}
}
int main(){scanf("%d",&N);for(int i=1;i<=N;i++){scanf("%d",&a[i]);b[i] = a[i];c[i] = a[i];}for(int i=1;i<=N;i++){scanf("%d",&res[i]);}insertionsort(b,res);heapsort(c,res);return 0;
}

1099 Build A Binary Search Tree (30 分)

算法标签: 树，二叉树 + 遍历
注意点: 先建树，然后执行中序遍历和层序遍历

#include<bits/stdc++.h>
using namespace std;
int N;
int a[105];struct node{int data;int lchild=-1;int rchild=-1;
};
node tree[105];
int cnt = 0;void InOrder(int root){if(root==-1){return;}InOrder(tree[root].lchild);tree[root].data = a[cnt++];InOrder(tree[root].rchild);
}
int res = 0;
void LevelOrder(int root){queue<int> q;q.push(root);while(!q.empty()){int f = q.front();q.pop();res++;if(res ==1){printf("%d",tree[f].data);}else{printf(" %d",tree[f].data);}if(tree[f].lchild!=-1){q.push(tree[f].lchild);}if(tree[f].rchild!=-1){q.push(tree[f].rchild);}}printf("\n");
}int main(){cin >> N;for(int i=0;i<N;i++){int l,r;cin >> l >> r;if(l==-1){tree[i].lchild = -1;}else{tree[i].lchild = l;}if(r==-1){tree[i].rchild = -1;}else{tree[i].rchild = r;}}   for(int i=0;i<N;i++){cin >> a[i];}sort(a,a+N);InOrder(0);LevelOrder(0);return 0;
}

1100 Mars Numbers (20 分)

算法标签: 模拟 + 字符串

#include<bits/stdc++.h>
using namespace std;
string D[13] = {"tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
string G[13] = {"","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
int isdigit(char c){if(c>='0' && c<='9'){return 1;}else{return 0;}
}
int iscase(char c){if(c>='a' && c<='z'){return 1;}else{return 0;}
}
int main(){int N;cin >> N;string s;getline(cin,s);for(int i=0;i<N;i++){string str;getline(cin,str);int num = 0;if(isdigit(str[0])){for(int i=0;i<str.size();i++){num = num *10 + str[i]-'0';}int g = num / 13;int d = num % 13;if(g!=0 && d==0){cout << G[g] << endl;}else if(g!=0 && d!=0){cout << G[g] << " "; cout << D[d] << endl;}else{cout << D[d] << endl;}}else{string s1 = "";string s2 = "";bool f1 = true;for(int i=0;i<str.size();i++){if(iscase(str[i]) && f1){s1 += str[i];}else if(str[i] == ' '){f1 = false;}else if(iscase(str[i])&& !f1){s2 += str[i];}}if(s2 == ""){for(int i=0;i<13;i++){if(s1 == D[i]){cout << i << endl;break;}}for(int i=1;i<13;i++){if(s1 == G[i]){cout << i*13 << endl;break;}}}else{int g,d;for(int i=1;i<13;i++){if(s1 == G[i]){g = i;break;}}for(int i=0;i<13;i++){if(s2 == D[i]){d = i;break;}}cout << g*13 + d << endl;}}}return 0;
}

PAT (Advanced Level) Practice 题解代码 - II （1051-1100）相关推荐

PAT (Advanced Level) Practice 1043 Is It a Binary Search Tree (25 分) 凌宸1642
PAT (Advanced Level) Practice 1043 Is It a Binary Search Tree (25 分) 凌宸1642 题目描述: A Binary Search Tr ...
PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) 凌宸1642
PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) 凌宸1642 题目描述: With the 2010 FIFA World Cu ...
【PAT (Advanced Level) Practice】1051 Pop Sequence (25 分)
1051 Pop Sequence (25 分) Given a stack which can keep M numbers at most. Push N numbers in the order ...
PAT (Advanced Level) Practice 题目集合（1001 ~ 1050）（正在更新）
1001 A+B Format (20 分) 题目大意:计算a+b,结果按照西方的那种写数字的方式输出,从三个数一个逗号那种. #include<bits/stdc++.h> using ...
【PAT (Advanced Level) Practice】1037 Magic Coupon (25 分)
题意: 给出两个集合,从这两个集合里面选出数量相同的元素进行一对一相乘,求能够得到的最大乘积之和. 题解: 对每个集合,将正数和负数分开考虑,将每个集合里的整数从大到小排序:将每个集合里的负数从小到大 ...
【PAT (Advanced Level) Practice】1093 Count PAT‘s (25 分)
1093 Count PAT's (25 分) The string APPAPT contains two PAT's as substrings. The first one is formed ...
【PAT (Advanced Level) Practice】1124 Raffle for Weibo Followers (20 分)
1124 Raffle for Weibo Followers (20 分) John got a full mark on PAT. He was so happy that he decided ...
【PTA】PAT (Advanced Level) Practice 1011-1014
目录 1011 World Cup Betting 翻译: 代码:(查找元素) 1012 The Best Rank 翻译: 代码:(排序) 1013 Battle Over Cities 翻译: 代 ...
PAT (Advanced Level) Practice - 1107 Social Clusters（30 分）
题目链接:点击打开链接题目大意:一个"社交集群"是指部分兴趣爱好相同的人的集合.你需要找出所有的社交集群. 解题思路:并查集思路,轮到第几个人时,以它的 id 为 root,然后 ...

PAT (Advanced Level) Practice 题解代码 - II （1051-1100）

PAT

PAT (Advanced Level) Practice - II（1051-1100）

1001-1167（已更新至最新！）

1051-1100

更多详见>>

题目传送门

序

1051 Pop Sequence (25 分)

1052 Linked List Sorting (25 分)

1053 Path of Equal Weight (30 分)

1054 The Dominant Color (20 分)

1055 The World’s Richest (25 分)

1056 Mice and Rice (25 分)

1057 Stack (30 分)

1058 A+B in Hogwarts (20 分)

1059 Prime Factors (25 分)

1060 Are They Equal (25 分)

1061 Dating (20 分)

1062 Talent and Virtue (25 分)

1063 Set Similarity (25 分)

1064 Complete Binary Search Tree (30 分)

1065 A+B and C (64bit) (20 分)

1066 Root of AVL Tree (25 分)

1067 Sort with Swap(0, i) (25 分)

1068 Find More Coins (30 分)

1069 The Black Hole of Numbers (20 分)

1070 Mooncake (25 分)

1071 Speech Patterns (25 分)

1072 Gas Station (30 分)

1073 Scientific Notation (20 分)

1074 Reversing Linked List (25 分)

1075 PAT Judge (25 分)

1076 Forwards on Weibo (30 分)

1077 Kuchiguse (20 分)

1078 Hashing (25 分)

1079 Total Sales of Supply Chain (25 分)

1080 Graduate Admission (30 分)

1081 Rational Sum (20 分)

1082 Read Number in Chinese (25 分)

1083 List Grades (25 分)

1084 Broken Keyboard (20 分)

1085 Perfect Sequence (25 分)

1086 Tree Traversals Again (25 分)

1087 All Roads Lead to Rome (30 分)

1088 Rational Arithmetic (20 分)

1089 Insert or Merge (25 分)

1090 Highest Price in Supply Chain (25 分)

1091 Acute Stroke (30 分)

1092 To Buy or Not to Buy (20 分)

1093 Count PAT’s (25 分)

1094 The Largest Generation (25 分)

1095 Cars on Campus (30 分)

1096 Consecutive Factors (20 分)

1097 Deduplication on a Linked List (25 分)

1098 Insertion or Heap Sort (25 分)

1099 Build A Binary Search Tree (30 分)

1100 Mars Numbers (20 分)

PAT (Advanced Level) Practice 题解代码 - II （1051-1100）相关推荐

最新文章

热门文章