POJ2151

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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 20.9 0.91 0.90 0 0

Sample Output

0.972

题意：有t支队伍，m道题,冠军最少做n道题，问保证每队最少做一题，冠军最少做n题的概率

思路：高中知识还真是没剩下多少了，下面转载别人博客中的解释，很详细，基本上看着这个思路，将之代码化就能过，注意精度。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>int m,n,t;double  dp[1010][40][40],s[1010][40],p[1010][40];using namespace std; int main(){while(~scanf("%d %d %d",&m,&t,&n) && (n || m || t)){memset(dp,0,sizeof(dp));memset(s,0,sizeof(s));for(int i = 1; i <= t; i++)for(int j = 1; j <= m; j++)scanf("%lf",&p[i][j]);for(int i = 1; i <= t; i++){dp[i][0][0] = 1.0;for(int j = 1; j <= m; j++)dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);for(int j = 1; j <= m; j++)for(int 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(int  j = 1; j <= m; j++)s[i][j] = s[i][j-1] + dp[i][m][j];}double p1,p2;p1 = p2 = 1.0;for(int i = 1; i <= t; i++)p1 *= (s[i][m] - s[i][0]);for(int i = 1; i <= t; i++) p2 *= (s[i][n-1] - s[i][0]);printf("%.3lf\n",p1-p2);}return 0;}

                                             
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