# POJ 2151 Check the difficulty of problems （概率DP）

Check the difficulty of problems

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 7304 Accepted: 3152

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

### 题意

ACM比赛中，共 M 道题，T 个队， p[i][j] 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。

### 思路

• ans: 所有队至少解出一题的概率
• as : 所有队至少解出一题但是不超过 n−1 题的概率

pd[j][k] 表示在前 j 道题中做对 k 道的概率，于是有

pd[j][k]=pd[j−1][k]∗(1.0−p[i][j])+pd[j−1][k−1]∗p[i][j]

### AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<stack>

double pd[35][35],p[1005][35]; //pd[i][j]表示在前i道题中做对j道的概率
int main()
{
int m,t,n;
while(~scanf("%d%d%d",&m,&t,&n)&&(m||t||n))
{
double ans=1.0,as=1.0;  //ans 为所有队至少A一道题的概率 as为所有队A的题目数量都不超过n的概率
memset(pd,0,sizeof(pd));
for(int i=0; i<t; i++)
{
double temp=1.0;
for(int j=1; j<=m; j++)
{
scanf("%lf",&p[i][j]);
temp*=1.0-p[i][j];  //逆命题就是该队伍一道题也没有做对
}
ans*=1.0-temp;      //因为要保证所有队伍都满足，所以用乘法定理
}
for(int i=0; i<t; i++)
{
memset(pd,0,sizeof(pd));
pd[0][0]=1.0;         //在前0道题目中A0道永真
for(int j=1; j<=m; j++)
{
pd[j][0]=pd[j-1][0]*(1-p[i][j]);
for(int k=1; k<=j; k++)
pd[j][k]=pd[j-1][k]*(1.0-p[i][j])+pd[j-1][k-1]*p[i][j]; //第j题要么做对，要么做错
}
double temp=0.0;
for(int j=1; j<n; j++)
temp+=pd[m][j];     //前m道题中做对数目少于n的子命题（做对1、2...n-1道）之间用加法定理
as*=temp;           //各组之间乘法定理
}
printf("%.3f\n",ans-as);
}
return 0;
}