http://poj.org/problem?id=2151   

                                                           Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4873		Accepted: 2131

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?

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.

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

2 2 2

0.9 0.9

1 0.9

0 0 0

0.972
分析：求保证每个队至少做对一题，冠军队做对n个题的概率。

保证每个队至少做对一题，冠军队做对n个题的概率=每个队至少做对一道题-没有一个队做到n到题。（每个队最多做了n-1个题），
dp[i][j][k]表示第i个对做到j题，目前做对了k题。
dp[i][j[k]=dp[i][j-1][k]*(1-a[i][j])+dp[i][j-1][k-1]*a[i][j];
s[i][k]表示i对至少做对了k题的概率
注意边界。

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

double dp[][][];

int main()

{

   int m,t,n,i,j,k;

   double a[][],cnt,ans,sum,s[][];

   while(~scanf("%d%d%d",&m,&t,&n))

   {

       memset(dp,,sizeof(dp));

       memset(s,,sizeof(s));

               cnt=;

               ans=;

               sum=;

       if(m==&&t==&&n==)

              break;

        for(i=;i<=t;i++)

          {

              for(j=;j<=m;j++)

              {

                 scanf("%lf",&a[i][j]) ;

                 cnt*=(-a[i][j]);

              }

              ans*=(-cnt);

              cnt=;

          }

           for(i=;i<=t;i++)

          {

             dp[i][][]=-a[i][];

              dp[i][][]=a[i][];

             for(j=;j<=m;j++)

                 dp[i][j][]=dp[i][j-][]*(-a[i][j]);

             for(j=;j<=m;j++)

              {

                 for(k=;k<=j;k++)

                 {

                 dp[i][j][k]=dp[i][j-][k]*(-a[i][j])+dp[i][j-][k-]*a[i][j];

                 }

              }

              for(k=;k<=n-;k++)

                 s[i][n-]+=dp[i][m][k];

             }

          for(i=;i<=t;i++)

          {

             sum*=s[i][n-];

          }

      printf("%.3lf\n",ans-sum);

   }

   return ;

}