标签:show mat 删掉 i++ 时间 gcd tle clu har
题意:给n个时间段(不会有交叉),在总时间L下,有多少个a的时间段可以抽烟。
#include <bits/stdc++.h> using namespace std; struct Line { int l,r; bool operator < (const Line & rhs) const { return r < rhs.r; } }; int main() { int n,l,a; vector<Line> v; //freopen("in.txt","r",stdin); scanf("%d%d%d",&n,&l,&a); for(int i = 0; i < n; i++) { int l,t; scanf("%d%d",&l,&t); v.push_back(Line{l,l+t}); } sort(v.begin(),v.end()); int ans = 0; for(int i = 0; i < (int)v.size()-1; i++) { ans += (v[i+1].l - v[i].r)/a; } if(n>0) { if(l>v[n-1].r) { ans+=(l-v[n-1].r)/a; } if(v[0].l>0) ans+=v[0].l/a; } else { ans += l/a; } cout<<ans<<endl; return 0; }
#include <bits/stdc++.h> using namespace std; const int maxn = 1000+5; char maps[maxn][maxn]; int n,m; int dx[8] = {-1,-1,-1,0,0,1,1,1}; int dy[8] = {-1,0,1,-1,1,-1,0,1}; bool inside(int x,int y) { bool flag = true; if(x>=1&&x<(n-1)&&y>=1&&y<(m-1)) { for(int i = 0; i < 8; i++) { int dr = x + dx[i]; int dc = y + dy[i]; if(maps[dr][dc]==‘.‘) flag = false; } } else flag = false; return flag; } bool calc(int i,int j) { if(inside(i-1,j-1)) return true; if(inside(i-1,j)) return true; if(inside(i-1,j+1)) return true; if(inside(i,j-1)) return true; if(inside(i,j+1)) return true; if(inside(i+1,j-1)) return true; if(inside(i+1,j)) return true; if(inside(i+1,j+1)) return true; return false; } int main() { //freopen("in.txt","r",stdin); scanf("%d%d",&n,&m); for(int i = 0; i < n; i++) scanf("%s",maps[i]); bool flag = true; for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { if(maps[i][j]==‘#‘) { if(!calc(i,j)) { flag = false; break; } } } if(!flag) break; } printf("%s\n",flag ? "YES":"NO"); return 0; }
#include <bits/stdc++.h> using namespace std; int main() { int n; scanf("%d",&n); int res = 1; while(n>3) { for(int i = 1; i <= (n+1)/2; i++) printf("%d ",res); res*=2; n/=2; } if(n==1) printf("%d\n",res); else if(n==2) printf("%d %d\n",res,res*2); else if(n==3) printf("%d %d %d\n",res,res,res*3); return 0; }
Codeforces Round #514 (Div. 2)
标签:show mat 删掉 i++ 时间 gcd tle clu har
原文地址:https://www.cnblogs.com/TreeDream/p/9765257.html