Check the difficulty of problems 概率dp，概率知识很重要

Problem 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

***************************************************************************************************************************

概率dp

***************************************************************************************************************************

/*
dp[i][j][k]代表第i个队在前j道题时解决k道题时的概率
p[i][j]代表第i个队解决第j道题时的概率
s[i][j]代表第i个队解决0-j道题的概率
注意要初始化dp[i][0][0]=1;
因为这是必然事件
要求有t个队，m道题，每队至少一题，且冠军队不少于n个题的事件概率
此时对冠军队不少于n个题，求对立事p2更容易，
每队至少一题直接求p1，两者相减就是要求的概率。
这道题说明了概率很重要。
*/
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
double dp[1005][50][50],p[1005][50],s[1005][50];
int n,m,t;
int i,j,k;
double p1,p2;
int main()
{
  while(scanf("%d%d%d",&m,&t,&n)!=EOF)
  {
      if(!m&&!n&&!t)
        break;
      memset(dp,0,sizeof(dp));
      memset(p,0,sizeof(p));
      memset(s,0,sizeof(s));
      for(i=1;i<=t;i++)
       for(j=1;j<=m;j++)
         scanf("%lf",&p[i][j]);
     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(j=1;j<=m;j++)
          s[i][j]=s[i][j-1]+dp[i][m][j];
     }
     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
",p1-p2);

  }
  return 0;
}