Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4091		Accepted: 1811

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?

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.

For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 using namespace std;

 int M,T,N;

 double a[][][],s[][],p1,pn,solve[][];

 int main()

 {

     while(~scanf("%d%d%d",&M,&T,&N))

     {

         if((M||T||N)==) break;

         for(int i=;i<=T;i++) for(int j=;j<=M;j++) scanf("%lf",&solve[i][j]);

         memset(a,,sizeof(a)); memset(s,,sizeof(s));

         for(int i=;i<=T;i++)

         {

             a[i][][]=;

             for(int j=;j<=M;j++)

             {

                 a[i][j][]=a[i][j-][]*(-solve[i][j]);

             }

         }

         for(int i=;i<=T;i++)

         {

             for(int j=;j<=M;j++)

             {

                 for(int k=;k<=j;k++)

                 {

                     a[i][j][k]=a[i][j-][k-]*solve[i][j]+a[i][j-][k]*(-solve[i][j]);

                 }

             }

         }

         for(int i=;i<=T;i++)

         {

             s[i][]=a[i][M][];

             for(int j=;j<=M;j++)

             {

                 s[i][j]=s[i][j-]+a[i][M][j];

             }

         }

         p1=pn=.;

         for(int i=;i<=T;i++)

         {

             p1*=s[i][M]-s[i][];

             pn*=s[i][N-]-s[i][];

         }

         printf("%.3lf\n",p1-pn);

     }

     return ;

 }