CodeForces 621A Wet Shark and Odd and Even】的更多相关文章

水题 #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #include<vector> #include<algorithm> using namespace std; int n; +]; int main() { scanf("%d",&n); ;i<n;i++) scanf("%lld",&…
水 最大偶数和 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack> #include<cmath…
A. Wet Shark and Odd and Even time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get ma…
题 Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the n …
http://codeforces.com/problemset/problem/621/E E. Wet Shark and Blocks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are b blocks of digits. Each one consisting of the same n digit…
B. Wet Shark and Bishops time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from…
E. Wet Shark and Blocks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are b blocks of digits. Each one consisting of the same n digits, which are given to you in the input. Wet Sha…
题意是得到最大的偶数和 解决办法很简单 排个序 取和 如果是奇数就减去最小的奇数 #include <cstdio> #include <cmath> #include <cstring> #include <queue> #include <vector> #include <algorithm> #define INF 0x3f3f3f3f #define mem(str,x) memset(str,(x),sizeof(str)…
方法可以转化一下,先计算每一个鲨鱼在自己范围内的数能被所给素数整除的个数有几个,从而得到能被整除的概率,设为f1,不能被整除的概率设为f2. 然后计算每相邻两只鲨鱼能获得钱的期望概率,f=w[id1].f1*w[id2].f2+w[id1].f2*w[id2].f1+w[id1].f1*w[id2].f1; f*2000就是这两只鲨鱼能获得的期望金钱,然后枚举一下所有相邻的鲨鱼,累加即可. #include<cstdio> #include<cstring> #include<…
记录一下每个对角线上有几个,然后就可以算了 #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #include<vector> #include<algorithm> using namespace std; +; int n; long long w1[maxn]; long long w2[maxn]; long long ans; int mai…