Check the difficulty of problems POJ 2151-优快云博客

本文链接：https://blog.youkuaiyun.com/JSPang/article/details/47948005

本文探讨了在编程竞赛中确保所有队伍至少解决一道题目，同时冠军队伍至少解决指定数量题目概率的计算方法。

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Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5862		Accepted: 2555

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

<span style="color:#cc0000;">题意：有t支队伍，m道题,冠军最少做n道题，问保证每队最少做一题，冠军最少做n题的概率</span>

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int m,t,n;
double dp[1005][35][35],s[1005][35],p[1005][35];
int main()
{
    int i,j,k;
    double p1,p2;
    while(scanf("%d%d%d",&m,&t,&n),m,t,n)
    {
        memset(dp,0,sizeof(dp));
        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++)
        {
            dp[i][0][0]=1.0;
            for(j=1;j<=m;j++)
               dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);
            for(j=1;j<=m;j++)
               for(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(k=1;k<=m;k++)
              s[i][k]=s[i][k-1]+dp[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;
}