hdu2151 Check the difficulty of problems

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4075		Accepted: 1804

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

主要是推出公式

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

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
double dp[1005][31][31],p[1005][31],sum[1005];
int main()
{
    int m,t,n,i,j,k;
    double ans1,ans2,temp;
    while(scanf("%d%d%d",&m,&t,&n)!=EOF&&m+t+n){
        for(ans1=1,i=0;i<t;i++){
            temp=1;
            for(j=0;j<m;j++){
                scanf("%lf",&p[i][j]);
                temp*=(1-p[i][j]);
            }
            ans1*=1.0-temp;
        }
        for(i=0;i<t;i++){
            for(j=0;j<m;j++){
                if(j)
                dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);
                else
                dp[i][j][0]=1-p[i][j];
            }
            dp[i][0][1]=p[i][0];
        }
        for(i=0;i<t;i++){
            for(j=1;j<m;j++){
                for(k=1;k<=j+1;k++)
                dp[i][j][k]=dp[i][j-1][k]*(1-p[i][j])+dp[i][j-1][k-1]*p[i][j];
            }
        }
        for(ans2=1,i=0;i<t;i++){
            sum[i]=0;
            for(k=1;k<n;k++){
                sum[i]+=dp[i][m-1][k];
            }
            ans2*=sum[i];
        }
        printf("%.3f\n",ans1-ans2);
    }
    return 0;
}