Educational Codeforces Round 88 (Rated for Div. 2)
题目链接:https://codeforces.com/contest/1359
题目讲解:https://www.bilibili.com/video/bv12z411v7WN
代码:
A.Berland Poker
// // main.cpp // CF // // Created by HanJinyu on 2020/5/15. // Copyright © 2020 by HanJinyu. All rights reserved. // #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <list> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; typedef double db; typedef long long ll; const int maxn=2e5+10; int main() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); #endif int t; scanf("%d",&t); while(t--) { int n,m,k; scanf("%d%d%d",&n,&m,&k); if(n/k>=m) { printf("%d\n",m); } else{ if((m-n/k)>=(k-1)) { int re=(m-n/k)/(k-1); if((m-n/k)%(k-1)!=0) re++; printf("%d\n",n/k-re); } else { printf("%d\n",n/k-1); } } } return 0; }
B.New Theatre Square
//------------------------------------------------- //Created by HanJinyu //Created Time :三 5/20 12:29:04 2020 //File Name :question.cpp //------------------------------------------------- #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <list> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; typedef double db; typedef long long ll; const int maxn = 2e5+10; int main() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); #endif int t; scanf("%d",&t); while(t--) { int n,m,x,y; scanf("%d%d%d%d",&n,&m,&x,&y); getchar(); char str[2000][2000]; int dian=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { cin>>str[i][j]; if(str[i][j]=='.') dian++; } } if(2*x<=y||m==1) { ll res=dian*x; printf("%lld\n",res); continue; } else if(n==1&&m==1) { printf("%d\n",x*dian); continue; } else if(n==1) { ll res1=0; bool used2[maxn]={false}; for(int j=0;j<m-1;j++) { if(str[0][j]=='.'&&str[0][j+1]=='.'&&!used2[j]) { res1+=y; used2[j]=true; used2[j+1]=true; } if(str[0][j]=='.'&&!used2[j]) res1+=x; } if(str[0][m-1]=='.'&&!used2[m-1]) res1+=x; printf("%lld\n",res1); continue; } else{ ll res=0; for(int i=0;i<n;i++) { bool used[maxn]={false}; for(int j=0;j<m-1;j++) { if(str[i][j]=='.'&&str[i][j+1]=='.'&&!used[j]) { used[j]=true; used[j+1]=true; res+=y; } else if(str[i][j]=='.'&&!used[j]) { res+=x; } } if(str[i][m-1]=='.'&&!used[m-1]) res+=x; } printf("%lld\n",res); } } return 0; }
C.Mixing Water
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <list> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; typedef double db; typedef long long ll; const int maxn=2e5+10; struct cha{ int a; int b; double chaa; bool operator<(const cha &a)const { return (this->chaa)<a.chaa; } }; int main() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); #endif int t; scanf("%d",&t); while(t--) { double h,c,t; scanf("%lf%lf%lf",&h,&c,&t); if(t>=h) printf("1\n"); else if(t<=(h+c)/2.0) printf("2\n"); else { int k=(t-h)/(h+c-2*t); //k个冷,k+1热,2*k+1 //k+1冷,k+2热,2*k+3 double cha1=((k+1)*h+k*c)/(2*k+1); double cha2=((k+2)*h+(k+1)*c)/(2*k+3); if(abs(cha1-t)<=abs(cha2-t)) printf("%d\n",2*k+1); else printf("%d\n",2*k+3); } } return 0; }
D.Yet Another Yet Another Task
// // main.cpp // CF // // Created by HanJinyu on 2020/5/15. // Copyright © 2020 by HanJinyu. All rights reserved. // #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <list> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; typedef double db; typedef long long ll; const int maxn=2e5+10; int main() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); #endif int n; scanf("%d",&n); int a[maxn]; for(int i=0;i<n;i++) scanf("%d",a+i); int ans=0; for(int i=1;i<=30;i++) { int cnt=0; for(int j=0;j<n;j++) { if(a[j]>i) cnt=0; else { cnt+=a[j]; if(cnt<0) cnt=0; else ans=max(ans,cnt-i); } } } printf("%d\n",ans); return 0; }