长春理工大学第十四届程序设计竞赛

Problem A Rubbish

https://ac.nowcoder.com/acm/contest/912/A

题意：1e5*1e5的棋盘上有n点，求连通块数量

C++版本一

题解：坐标离散化+坐标数轴化+二分+递推+并查集

1、对所有坐标离散化成数轴；

2、由左上到右下递推；

3、并查集合并；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const ll INF = 0x3f3f3f3f3f3f3f;
int s,t,n,m,k,p,l,r;
ll u,v,a[N];
int x[N],y[N],b[N];
ll ans,cnt,flag,temp,sum;
int pre[N];
int find(int x){if(x==pre[x])return x;return pre[x]=find(pre[x]);
}
void marge(int x,int y){int tx=find(x);int ty=find(y);if(tx!=ty){pre[tx]=ty;}
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//int T=0;while(~scanf("%d",&n)){for(int i=1;i<=n;i++)pre[i]=i;for(int i=1;i<=n;i++){scanf("%d%d",&x[i],&y[i]);a[i]=(x[i]-1)*100000ll+y[i];}sort(a+1,a+1+n);for(int i=1;i<=n;i++){u=x[i];v=y[i];ll now=(u-1)*100000ll+v;int ii=lower_bound(a+1,a+n+1,now)-a;if(u<100000ll&&v<100000ll){int jj=lower_bound(a+1,a+n+1,u*100000ll+v+1)-a;if(jj<=n&&a[jj]==u*100000ll+v+1)marge(ii,jj);}if(v<100000ll){int ij=lower_bound(a+1,a+n+1,(u-1)*100000ll+v+1)-a;if(ij<=n&&a[ij]==(u-1)*100000ll+v+1)marge(ii,ij);}if(u<100000ll){int ji=lower_bound(a+1,a+n+1,u*100000ll+v)-a;if(ji<=n&&a[ji]==u*100000ll+v)marge(ii,ji);}if(u<100000ll&&v>1){int jii=lower_bound(a+1,a+n+1,u*100000ll+v-1)-a;if(jii<=n&&a[jii]==u*100000ll+v-1)marge(ii,jii);}if(u>1&&v<100000ll){int jji=lower_bound(a+1,a+n+1,(u-2)*100000ll+v+1)-a;if(jji<=n&&a[jji]==(u-2)*100000ll+v+1)marge(ii,jji);}}ans=0;for(int i=1;i<=n;i++)ans+=(pre[i]==i);cout<<ans<<endl;//printf("%lld\n",ans);}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}
/*
4
1 2
2 1
2 3
3 2
*/

C++版本二

题解：坐标离散化+并查集

将格点按行先列次排序，遍历过程更新当前列的格点中行数最大格点的下标，判断能否与当前点连通，用并查集维护连通块。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define PB push_back
#define endl '\n'
const int N=1e6+7;
const int INF=1e8,mod=1e7+7;
int n,m,k;
struct s{int x,y;int fa;
}a[N];
int f[N];
int F(int x){return x==a[x].fa?x:a[x].fa=F(a[x].fa);
}
void u(int x,int y){x=F(x),y=F(y);if(x==y)return;a[x].fa=y;
}
bool cmp(s p,s q){if(p.x==q.x)return p.y<q.y;return p.x<q.x;
}
int main()
{cin>>n;for(int i=1;i<=n;i++){scanf("%d%d",&a[i].x,&a[i].y);}memset(f,-1,sizeof(f));sort(a+1,a+1+n,cmp);for(int i=1;i<=n;i++)a[i].fa=i;a[0].y=a[0].x=-1;for(int i=1;i<=n;i++){//如果与前一个格点同行且下标差一说明连通if(a[i].y==a[i-1].y+1&&a[i].x==a[i-1].x){u(a[i].fa,a[i-1].fa);}//左上是否有格点if(a[f[a[i].y-1]].x==a[i].x-1){u(a[i].fa,a[f[a[i].y-1]].fa);}//上方是否有if(a[f[a[i].y]].x==a[i].x-1){u(a[i].fa,a[f[a[i].y]].fa);}//右上是否有if(a[f[a[i].y+1]].x==a[i].x-1){u(a[i].fa,a[f[a[i].y+1]].fa);}f[a[i].y]=i;}memset(f,0,sizeof(f));int ans=0;for(int i=1;i<=n;i++){if(!f[F(a[i].fa)]){ans++;f[F(a[i].fa)]=1;}//cout<<i<<' '<<a[i].x<<' '<<a[i].y<<' '<<a[i].fa<<' '<<ans<<endl;}cout<<ans;return 0;
}

Problem B Bowling Game

https://ac.nowcoder.com/acm/contest/912/B

题意：蓝色面积S1，黄色面积S2，问球的直径多大的时候会按照图中所示卡住。

C++版本一

题解：几何数学

设d为直径，r为半径，S2的边分别为a，b，c，其中c*c=S1；

$d=2*r=2*2*\frac{s_{1}}{\sqrt{s_{2}}+\sqrt{s_{2}+4*s_{1}}}$

证明：

$\because \left\{\begin{matrix} s_{1}=\frac{1}{2}*a*b \\s_{2}=c^{2} \\a^{2}+b^{2}=c^{2} \end{matrix}\right.$

$\therefore a+b+c=\sqrt{s_{2}}+\sqrt{s_{2}+4*s_{1}}$

$\because \frac{1}{2}*(a+b+c)*r=s_{1}$

$\therefore r=2*\frac{s_{1}}{\sqrt{s_{2}}+\sqrt{s_{2}+4*s_{1}}}$

$\because d=2*r$

$\therefore d=2*r=2*2*\frac{s_{1}}{\sqrt{s_{2}}+\sqrt{s_{2}+4*s_{1}}}$

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%lld%lld",&n,&m);printf("%f\n",4*n/(sqrt(m)+sqrt(m+4*n)));//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

C++版本二

题解：二分

对于一个斜边确定的等腰三角形，两直角边差越小面积越大，用s1求出斜边，然后二分法求两直角边的长度，内切圆直径就是直角边长和减斜边长。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define PB push_back
#define endl '\n'
int main()
{double s1,s2;cin>>s1>>s2;double x=sqrt(s2);double l=0,r=sqrt(2.0)/2*x;while(r-l>=0.0000001){double mid=(l+r)/2;if(mid*sqrt(x*x-mid*mid)/2>s1){r=mid;}else l=mid;}double ans=(l+sqrt(x*x-l*l)-x);printf("%.6lf",ans);return 0;
}

Problem C REN

https://ac.nowcoder.com/acm/contest/912/C

题意：

题解：

C++版本一

Problem D Capture The Flag

https://ac.nowcoder.com/acm/contest/912/D

题意：

题解：

C++版本一

Problem E Shortest Code

https://ac.nowcoder.com/acm/contest/912/E

题意：去掉没必要的空格、空行和注释。

题解：模拟

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
string a;
string res;
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){//scanf("%d",&n);while(getline(cin,a)){int l=a.size();res="";for(int i=0;i<l;i++)if(a[i]==' ')if(i!=0&&i!=l-1&&isalnum(a[i-1])&&isalnum(a[i+1]))res+=' ';else a[i]=a[i-1];//将空格更新为前一个值else if(i<l-1&&a[i]=='/'&&a[i+1]=='/')break;else res+=a[i];if(res.size())cout<<res<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem F Successione di Fixoracci

https://ac.nowcoder.com/acm/contest/912/F

题意：

已知a和b的值。现在你只要求出Tn是多少

题解：规律

C++版本一

题解：

1、任意一个数被同一个数异或两次等于本身。即a^b^b=a；

2、此数列为 a，b，a^b循环

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%lld%lld%lld",&n,&m,&k);if(k%3==0)ans=n;if(k%3==1)ans=m;if(k%3==2)ans=n^m;cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem G Segment Tree

https://ac.nowcoder.com/acm/contest/912/G

题意：

题解：

C++版本一

Problem H Arithmetic Sequence

https://ac.nowcoder.com/acm/contest/912/H

题意：输出一个等差数列，使得数列和等于X

题解：没有要求数列长度，因此直接输出x

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);cout<<1<<endl;cout<<n<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem I Fate Grand Order

https://ac.nowcoder.com/acm/contest/912/I

题意：最多n/3个活动，使得期望最大

题解：贪心

每张卡片带来开心值的期望为抽中概率乘开心值，按期望排序贪心即可

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,l,r,u,v;
int ans,cnt,flag,temp,sum;
string str="";
struct node{double p,x;double q;int id;bool operator <(const node&S)const{return q>S.q;}
}e[N];
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=m;i++)scanf("%lf",&e[i].p),e[i].id=i,str+='0';for(int i=1;i<=m;i++)scanf("%lf",&e[i].x),e[i].q=e[i].x*e[i].p;n/=3;sort(e+1,e+m+1);for(int i=1;i<=min(n,m);i++)str[e[i].id-1]='1';cout<<str<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem J Printout

https://ac.nowcoder.com/acm/contest/912/J

题意：

题解：模拟

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
double ans,cnt,flag,temp,sum;
int a[N];
double b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=m;i++)scanf("%d",&a[i]);for(int i=1;i<=m;i++)scanf("%lf",&b[i]);a[0]=1;for(int i=1;i<=m;i++){if(n<=a[i]){ans=n*b[i];for(int j=i+1;j<=m;j++){ans=min(ans,a[j-1]*b[j]);}break;}}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem K Master of Graph

https://ac.nowcoder.com/acm/contest/912/K

题意：区间修改+区间查询

题解：线段树+标记

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v,q;
int ans,cnt,flag,temp,sum;
char str;
struct node{};
ll tree[N<<2];
bool tag[N<<2];
ll f(ll x){//计算f(x)ll re=0;while(x){re+=x%10;x/=10;}return re;
}
void pushup(int rt){tree[rt]=tree[rt<<1]+tree[rt<<1|1];tag[rt]=tag[rt<<1]&&tag[rt<<1|1];
}
void build(int l,int r,int rt){if(l==r){scanf("%lld",&tree[rt]);tag[rt]=0;return;}int mid=(l+r)>>1;build(l,mid,rt<<1);build(mid+1,r,rt<<1|1);pushup(rt);}
void updata(int l,int r,int rt,int L,int R){if(tag[rt])return;if(l==r){tree[rt]=f(tree[rt]);if(tree[rt]<10)tag[rt]=1;return;}int mid=(l+r)>>1;if(L<=mid)updata(l,mid,rt<<1,L,R);if(R>mid)updata(mid+1,r,rt<<1|1,L,R);pushup(rt);
}
ll query(int l,int r,int rt,int L,int R){if(L<=l&&r<=R){return tree[rt];}int mid=(l+r)>>1;if(R<=mid)return query(l,mid,rt<<1,L,R);else if(L>mid)return query(mid+1,r,rt<<1|1,L,R);else return query(l,mid,rt<<1,L,R)+query(mid+1,r,rt<<1|1,L,R);
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);memset(tree,0,sizeof(tree));build(1,n,1);while(m--)scanf("%d%d",&u,&v);scanf("%d",&q);while(q--){scanf("%d%d%d",&t,&l,&r);if(t){updata(1,n,1,l,r);}else{cout<<query(1,n,1,l,r)<<endl;}}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem L Homework Stream

https://ac.nowcoder.com/acm/contest/912/L

题意：编号为k的作业依赖于哪些作业，以及哪些作业依赖于编号为k的作业。

题解：模拟

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
vector<int>a,b;
void print(vector<int>p){for(int i=0;i<p.size();i++){printf("%d%c",p[i]," \n"[i==p.size()-1]);}if(p.size()==0)cout<<endl;
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);if(u==k)a.push_back(v);if(v==k)b.push_back(u);}print(a);print(b);//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem M Orx Zone

https://ac.nowcoder.com/acm/contest/912/M

题意：

题解：对于1-n为左边界的情况恭喜就是(n+1-当前最近的’x’和’r’的下标的较大值）

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,x[N],r[N],u,v;
ll ans,cnt,flag,temp,sum;
char a[N];
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){//scanf("%d",&n);scanf("%s",a+1);int len=strlen(a+1);x[len+1]=r[len+1]=len+1;for(int i=len;i>=1;i--){x[i]=(a[i]=='x'?i:x[i+1]);//在该点后最近的xr[i]=(a[i]=='r'?i:r[i+1]);//在该点后最近的r}for(int i=1;i<=len;i++)ans+=len-max(x[i],r[i])+1;cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}