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

题意:有t个队伍,然后有m道题,给出第i个队伍做出第j道题的概率pij,问在所有的队伍至少ac一道题,冠军至少acN道题的概率。

思路:我们可以以dpi[k]表示第i个队伍前j题做出k道题的概率,然后前缀和优化一下,dpi[j]表示第i个队伍做出j题以内的概率。然后用所有队伍至少做出一道道概率减去所有队伍做题范围在1~n-1的概率,就是所求概率,不过按照这样话冠军是可以不止一个的。

#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;
}

 

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