Check the difficulty of problems POJ 2151

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5862 Accepted: 2555

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

Source

POJ Monthly,鲁小石
    

<span style="color:#cc0000;">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</span>

    

#include <iostream>#include<stdio.h>#include<string.h>using namespace std;int m,t,n;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),m,t,n)    {        memset(dp,0,sizeof(dp));        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(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;}