Check the difficulty of problems - poj 2151 (概率+DP)

有 T(1<T<=1000) 支队伍和 M(0<M<=30) 个题目，已知每支队伍 i 解决每道题目 j 的的概率 p[i][j]，现在问：每支队伍至少解决一道题，且解题最多的队伍的题目数量不少于 N(0<N<=M) 的概率是多少？

p[i][j]表示第i个队伍做对第j题的概率。dp[i][j][k]表示第i个队伍在前j题中做对了k道的概率。
dp[i][j][k] = dp[i][j-1][k-1]*(p[i][j])+dp[i][j-1][k]*(1-p[i][j]);
再求出每个队都至少做对 1 道题的概率：ans1 *= 1 - dp[i][m][0];
求出每个队都只做对了 1 ～ n-1 题的概率 ans2即：(把每个队做对 1 ～ n-1 题的概率相加后，并把每个队的结果相乘）；
然后两者相减ans1-ans2

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 int M,T,N;
 double p[][];
 double dp[][][];
 double sum[][];
 int main() {
     scanf("%d %d %d",&M,&T,&N);
     while(M&&N&&T){
         memset(p,,sizeof(p));
         for(int i=;i<=T;i++){
             for(int j=;j<=M;j++){
                 scanf("%lf",&p[i][j]);
             }
         }
         memset(dp,,sizeof(dp));
         for(int i=;i<=T;i++){
             dp[i][][]=p[i][];
             dp[i][][]=-p[i][];

         }
         for(int i=;i<=T;i++){
             for(int j=;j<=M;j++){
                 for(int k=;k<=j;k++){
                     dp[i][j][k]=dp[i][j-][k]*(1.0-p[i][j]);
                     if(k>) dp[i][j][k]+=dp[i][j-][k-]*p[i][j];
                 }
             }
         }
         memset(sum,,sizeof(sum));
         for(int i=;i<=T;i++){
              sum[i][]=dp[i][M][];
             for (int j=;j<=M;j++) {
                    sum[i][j]=sum[i][j-]+dp[i][M][j];
             }
         }
         double ans1=1.0,ans2=1.0;
         for(int i=; i<=T; i++)  {
                ans1*=sum[i][M]-sum[i][];
                ans2*=(sum[i][N-]-sum[i][]);
             }
         printf("%.3lf\n",ans1-ans2);
         scanf("%d %d %d",&M,&T,&N);
     }

     return ;
 }

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5766		Accepted: 2515

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