水题. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; struct X { int x,y; }s[],tmp[]; int n; int main() { scanf("%d",&n); ;i<=n;i++) scanf("%d%d",&s[i].x,&…

A. Wilbur and Swimming Pool Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/596/problem/A Description After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his back…

A. Wilbur and Swimming Pool time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in…

C time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set u…

 A - Wilbur and Swimming Pool Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 596A Description After making bad dives into swimming pools, Wilbur…

Description After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the a…

Description After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the a…

http://codeforces.com/contest/596/problem/D 题目大意: 有n棵树排成一排,高度都为h. 主人公要去砍树,每次等概率地随机选择没倒的树中最左边的树或者最右边的树把它砍倒.每棵树被砍到后,有p的概率会往左边倒,(1-p)的概率往右边倒. 树倒下后如果压到别的树,即如果那棵树倒下的方向上距离不到h的地方还有一棵树,,那么那棵树也会朝和这个树相同的方向倒下. 问最后所有的树都被砍完后覆盖的地面的长度的期望. 思路:dp[i][j][0/1][0/1]代表i到j…

一直在考虑, 每一段的贡献, 没想到这个东西能直接dp..因为所有的h都是一样的. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ…

题意:一个矩阵上面有0~9的数字,可以从任意一个格子出发,每次根据格子上的数字会前进到另一个格子(或原地不动),现在给出q个数位串,问是否有走法可以取出这个串(走到格子上的时候可以不取). 思路:发现每个点的出度至多为1,那么可以预处理出走到这个格子后取某个数字最近的格子是哪个(f[i][j][0-9]),然后用队列存下开头,在做一遍贪心就可以了 #include<cstdio> #include<cmath> #include<cstring> #include<…