题目链接: E. ZS and The Birthday Paradox. time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that giv
Polygon John has been given a segment of lenght N, however he needs a polygon. In order to create a polygonhe has cut given segment K times at random positions (uniformly distributed cuts). Now he has K + 1much sh
2020-02-10 13:24:26 问题描述: 问题求解: 第一次看到这个题目的时候想到的是之前做过的一条概率题,就是那个国际象棋,那条题目的做法是使用dp去对所有留在棋盘的种类进行计数,然后除以总数.所以,我一开始的做法也是这样做的. 但是实际是不对的,问题就是最后得到的解的概率是不等的,所以不能简单的计数符合条件的个数和最后的总个数并相除,如下面的例子. If you draw 1, then 9, probability is P(1) * P(9) = 0.1 * 0.1 = 0.0
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int N=1e2+10; int main() { int T,cas=1,n,x; scanf("%d",&T); while(T--) { scanf("%d",&n); int up=0; int down=n; for(int i=1;i<=n;i++) { scanf(&qu