Check the difficulty of problems poj 2151 dp

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7079 Accepted: 3066

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

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int mod=100000007;double dp[1005][31][31];double p[1005][31],s[1005][31];int m,t,n,i,j;void Clear(){      memset(s,0,sizeof(s));      memset(dp,0,sizeof(dp));      for(i=1;i<=t;i++)        dp[i][0][0]=1.0;      for(i=1;i<=t;i++)        for(j=1;j<=m;j++)        dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);}int main(){      int k;      while(~scanf("%d%d%d",&m,&t,&n))      {           if(m+t+n==0)            break;           for(i=1;i<=t;i++)            for(j=1;j<=m;j++)            scanf("%lf",&p[i][j]);           Clear();           for(i=1;i<=t;i++)           {               for(j=1;j<=m;j++)               {                   for(k=1;k<=m;k++)                    dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);               }           }           for(i=1;i<=t;i++)            for(k=1;k<=m;k++)            s[i][k]=s[i][k-1]+dp[i][m][k];           double p1=1,p2=1;           for(i=1;i<=t;i++)           {               p1*=s[i][m]-s[i][0];               p2*=s[i][n-1]-s[i][0];           }           printf("%.3lf\n",p1-p2);      }}