POJ 2151 Check the difficulty of problems

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4011		Accepted: 1779

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

Source

POJ Monthly,鲁小石 

设a[i][j][k]表示第i队在前j道题中共解出k道题的概率，易得a[i][j][k]有如下递推
关系(另需考虑边界条件)：
a[i][j][k] = a[i][j-1][k-1] * p[i][j] + a[i][j-1][k] * (1-p[i][j])
设s[i][j]表示a[i][M][0] + a[i][M][1] + ... + a[i][M][j]
问题的解可以转化为：每队均至少做一题的概率（用P1表示）减去每队做题数均在1到N-1
之间的概率（用P2表示）。
P1 = (s[1][M] - s[1][0])*(s[2][M]-s[2][0])*...*(s[T][M]-s[T][0])
P2 = (s[1][N-1] - s[1][0])*(s[2][N-1]-s[2][0])*...*(s[T][N-1]-s[T][0])

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

double dp[1100][40][40];
double p[1100][40];
double sum[1100][40];
double p1,pn;
int M,T,N;

int main()
{
    while(scanf("%d%d%d",&M,&T,&N)!=EOF)
    {
        if((M||T||N)==0) break;

        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        memset(p,0,sizeof(p));

        for(int i=1;i<=T;i++)
        {
            p[i][0]=1;
            for(int j=1;j<=M;j++)
            {
                scanf("%lf",&p[i][j]);
            }
        }

        for(int i=1;i<=T;i++)
        {
            dp[i][0][0]=1;
        }

        for(int i=1;i<=T;i++)
        {
            for(int j=1;j<=M;j++)
            {
                for(int k=0;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]);
                }
            }
        }

        for(int i=1;i<=T;i++)
        {
            for(int j=0;j<=M;j++)
            {
                for(int k=0;k<=j;k++)
                    sum[i][j]+=dp[i][M][k];
            }
        }
/*
        for(int i=1;i<=T;putchar(10),i++)
        {
            for(int j=1;j<=M;putchar(10),j++)
            {
                for(int k=0;k<=M;k++)
                {
                    printf("dp[%d][%d][%d]:   %lf
",i,j,k,dp[i][j][k]);
                }
            }
        }

        for(int i=1;i<=T;putchar(10),i++)
        {
            for(int k=0;k<=M;k++)
            {
                printf("sum[%d][%d]:  %lf
",i,k,sum[i][k]);
            }
        }
*/
        p1=1;pn=1;
        for(int i=1;i<=T;i++)
        {
            p1*=(sum[i][M]-sum[i][0]);
            pn*=(sum[i][N-1]-sum[i][0]);
        }
        printf("%.3lf
",p1-pn);
    }

    return 0;
}