摘要：
poj 2096详解见：http://blog.csdn.net/morgan_xww/article/details/6774708View Code #include<iostream>using namespace std;const int N = 2000;double dp[N][N];int n,s;double dfs(int x,int y){ if(x==n&&y==s) return 0; if(x>n||y>s) return 0; if(dp[x][y]!=0) return dp[x][y]; dp[x+1][y]=dfs(x 阅读全文
