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  • Collecting Bugs POJ2096

    题目链接

    简单的期望DP, 只不过是二维形式, 设(dp_{i,j})为两种分别选了(i, j)种后还需要的期望数, 则(dp_{i,j} = frac{i*j}{n*s}*dp_{i,j} + frac{(n-i)*j}{n*s}*dp_{i+1,j} + frac{i*(s-j)}{n*s}*dp_{i,j+1} + frac{(n-i)*(s-j)}{n*s}*dp_{i+1,j+1} + 1), 移项整理即可

    #include<iostream>
    #include<vector>
    #include<algorithm>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    
    const int maxn = 1007;
    double dp[maxn][maxn];
    
    
    void run_case() {
        int n, s;
        cin >> n >> s;
        for(int i = n; i >= 0; --i)
            for(int j = s; j >= 0; --j) {
                if(i == n && j == s) continue;
                dp[i][j] = (n*s +(n-i)*j*dp[i+1][j]+(s-j)*i*dp[i][j+1]+(n-i)*(s-j)*dp[i+1][j+1])/(n*s-i*j);
            }
        cout << dp[0][0];
    }
    
    
    int main() {
        //ios::sync_with_stdio(false), cout.tie(0);
        cout.flags(ios::fixed);cout.precision(4);
        run_case();
        //cout.flush();
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/GRedComeT/p/12452073.html
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