poj 2151 Check the difficulty of problems - dblank

# poj 2151 Check the difficulty of problems

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f, N = 1e3 + 10, Mod = 100000007;
typedef long long ll;
int n, m, t;
double pN, dpN[32];
int main()
{
while(cin>>m>>t>>n && m&&t&&n)
{
for(int i = 1; i<=t; i++)
for(int j = 1; j<=m; j++)
{
scanf("%lf", &pi);
}
memset(dp, 0, sizeof(dp));
dp0[0] = 1;
for(int i = 1; i<=t; i++)
{
dpi[0] = 1;
for(int  j = 1; j<= m; j++)
for(int k = 0; k<=j; k++)
{
dpi[k] = dpi[k] * (1.0 - pi);
if(k) dpi[k] += dpi[k-1] * pi;
}
}
for(int i = 1; i<=t; i++)
for(int j = 1; j<=m; j++){
dpi[j] += dpi[j-1];
}
double p1 = 1, p2 = 1;
for(int i = 1; i<=t; i++)
{
p1 *= dpi[m] - dpi[0];
p2 *= dpi[n-1] - dpi[0];
}
printf("%.3fn", p1 - p2);
}
return 0;
}