POJ2151：Check the difficulty of problems(概率DP)

Description

Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms:
1. All of the teams solve at least one problem.
2. The champion (One of those teams that solve the most problems) solves at least a certain number of problems.

Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem.

Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability Pij (1 <= i <= T, 1<= j <= M). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems?

Input

The input consists of several test cases. The first line of each test case contains three integers M (0 < M <= 30), T (1 < T <= 1000) and N (0 < N <= M). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just Pij. A test case of M = T = N = 0 indicates the end of input, and should not be processed.
Output

For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.
Sample Input

2 2 2
0.9 0.9
1 0.9
0 0 0
Sample Output

0.972 尼瑪，這道題卡精度上了，輸出要用單精度而不能雙精度，否則WA至死。題意：有t支隊伍，m道題,冠軍最少做n道題，問保證每隊最少做一題，冠軍最少做n題的概率思路：高中知識還真是沒剩下多少了，下面轉載別人博客中的解釋，很詳細，基本上看著這個思路，將之代碼化就能過，注意精度。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n,m,t;
double dp[1005][35][35],s[1005][35],p[1005][35];

int main()
{
    int i,j,k;
    double p1,p2;
    while(~scanf("%d%d%d",&m,&t,&n))
    {
        if(!m && !t && !n)
            break;
        for(i = 1; i<=t; i++)
            for(j = 1; j<=m; j++)
                scanf("%lf",&p[i][j]);
        memset(dp,0,sizeof(dp));
        memset(s,0,sizeof(s));
        for(i = 1; i<=t; i++)
        {
            dp[i][0][0] = 1.0;
            for(j = 1; j<=m; j++)
                dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);

            for(j = 1; j<=m; j++)
                for(k = 1; k<=j; k++)
                    dp[i][j][k] = dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);

            s[i][0] = dp[i][m][0];
            for(k = 1; k<=m; k++)
                s[i][k] = s[i][k-1]+dp[i][m][k];
        }
        p1 = p2 = 1.0;
        for(i = 1; i<=t; i++)
            p1*=(s[i][m]-s[i][0]);
        for(i = 1; i<=t; i++)
            p2*=(s[i][n-1]-s[i][0]);
        printf("%.3f\n",p1-p2);
    }


    return 0;
}