POJ2151-----Check the difficulty of problems

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本文深入探讨了在编程竞赛中组织者如何通过预赛结果估算参赛团队解决特定问题的概率，进而计算出所有参赛队伍至少解决一题且冠军团队至少解决指定数量题目概率的复杂过程。

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Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5178		Accepted: 2291

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

Sample Output

0.972

Source

POJ Monthly,鲁小石

首先要计算出每一个人做出一定题目的概率
令dp[i][j][k] 表示第i个人在前j题里ACk道题的概率
处理完以后，计算每一个人都至少AC一题的概率
在计算出每一个人做出的题目数都在[1, n] 的概率
两者之差就是答案，它们差的含义就是每一个人都至少做出一题，但不是所有人的出题数都在[1, n]

/*************************************************************************
    > File Name: POJ2151.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2014年12月24日 星期三 16时53分17秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

double dp[1010][35][35];
double p[1010][35];

int main()
{
	int m, t, n;
	while (~scanf("%d%d%d", &m, &t, &n))
	{
		if (!m && !t && !n)
		{
			break;
		}
		double x = 1.0, y = 1.0;
		memset (dp, 0, sizeof(dp));
		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)
			{
				double tmp = 1;
				for (int k = 1; k <= j; ++k)
				{
					tmp *= (1 - p[i][k]);
				}
				dp[i][j][0] = tmp;
				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]);
				}
			}
		}
		for (int i = 1; i <= t; ++i)
		{
			double tmp = 0;
			for (int j = 1; j <= m; ++j)
			{
				tmp += dp[i][m][j];
			}
			x *= tmp;//每一个队伍至少AC一道题的概率
		}
		for (int i = 1; i <= t; ++i)
		{
			double tmp = 0;
			for (int j = 1; j < n; ++j)
			{
				tmp += dp[i][m][j];
			}
			y *= tmp;
		}
		printf("%.3f\n", x - y);
	}
	return 0;
}