POJ2151 : Check the difficulty of problems

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

思路分析：

     因为是我们最后需要算各个队伍各自解决问题个数的概率

     我们开一个三维数组 F[i][j][k]表示第 i 支队伍在前 j 道题中共解决了 k 道题。

     然后开一个二维数组 S[i][k] 这个数组的作用在后面再说。

     首先我们清零F数组，我们可以很容易的知道这个数组中某些元素的值

     比如 F[i][0][0] = 1.0  第i支队伍在前0道题中解决0道题的概率肯定是1.0

     还有 F[i][j][0] = F[i][j-1][0] * ( 1 - P[i][j])

     前j道题中 做出0道题的概率 等于 前j-1道题中 做出 0 道题的概率 乘上 第j道题做不出来的概率

      把这些很显然的元素放进去之后，我们就得到了F的边界状态了。

     然后尝试写出状态转移方程：

       F[i][j][k] = F[i][j-1][k] * (1 - P[i][j]) + F[i][j-1][k-1] * P[i][j]

    然后我们看看S[i][k]这个数组中保存的是什么东西...

    先给出S[i][k]是怎么得到的

       S[i][k] = F[i][M][0] + F[i][M][1] + F[i][M][2] + ... + F[i][M][k]

    他的意思是：第i支队伍做出 不超过k道题 的概率  

    在所有题中做出0道，做出1道，做出2道，做出3道...做出k道。

    S[i][M] - S[i][0] 就表示第 i 支队伍做出 至少一道题 的概率

    同理 S[i][N-1]-S[i][0]就表示第 i 支队伍做出的题在 1 到 N-1 道之间的概率

    那么我们把所有队伍的这两个概率连乘，得到两个概率P1，P2。

    P1 表示 所有队伍至少做出一道题的概率

    P2 表示 所有队伍至少做出一道题，且至多做出N-1道题的概率

    P1 - P2 就表示 所有队伍至少做出一道题，且有队伍做出至少 N 道题的概率，也就是我们的计算目标。

代码如下：

#include <stdio.h>#include <stdlib.h>#include<string.h>double F[1010][35][35];double P[1010][35];double S[1010][35];int main(){    int M,T,N;    int i,j,k;    double P1,P2;    while(scanf("%d %d %d",&M,&T,&N) != EOF)    {        if(!M && !N && !T)            break;        memset(F,0,sizeof(F));        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++)        {            F[i][0][0] = 1.0;              for(j=1;j<=M;j++)               F[i][j][0] = F[i][j-1][0] * (1 - P[i][j]);              for(j=1;j<=M;j++)               for(k=1;k<=j;k++)                 F[i][j][k] = F[i][j-1][k]*(1-P[i][j]) + F[i][j-1][k-1] * P[i][j];            S[i][0] = F[i][M][0];            for(k=1;k<=M;k++)                         S[i][k] = S[i][k-1] + F[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;}