poj 2151 Check the difficulty of problems

题目链接：点击打开链接

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

题意：
ACM比赛中，共M道题，T个队，pij表示第i队解出第j题的概率
问 每队至少解出一题且冠军队至少解出N道题的概率。


解析：DP
设dp[i][j][k]表示第i个队在前j道题中解出k道的概率
则：
dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k]);
先初始化算出dp[i][0][0]和dp[i][j][0];
设s[i][k]表示第i队做出的题小于等于k的概率
则s[i][k]=dp[i][M][0]+dp[i][M][1]+``````+dp[i][M][k];


则每个队至少做出一道题概率为P1=(1-s[1][0])*(1-s[2][0])*```(1-s[T][0]);
每个队做出的题数都在1~N-1的概率为P2=(s[1][N-1]-s[1][0])*(s[2][N-1]-s[2][0])*```(s[T][N-1]-s[T][0]);


最后的答案就是P1-P2

<span style="font-size:24px;">#include <iostream>#include<cstdio>using namespace std;double dp[1111][50][50];double p[1111][50];double s[1111][50];int main(){    int m,t,n;    while(cin>>m>>t>>n)    {        if(m==0&&t==0&&n==0)break;        for(int i=1;i<=t;i++)        {            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 j=1;j<=m;j++)dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);            for(int j=1;j<=m;j++)            {                for(int k=1;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]);                }            }             s[i][0]=dp[i][m][0];             for(int k=1;k<=m;k++)s[i][k]=s[i][k-1]+dp[i][m][k];        }        double p1=1;        double p2=1;        for(int j=1;j<=t;j++)        {            p1=p1*(1-s[j][0]);            p2=p2*(s[j][n-1]-s[j][0]);        }        printf("%.3f\n",p1-p2);    }    return 0;}</span>