再减去每个队都只做对1～m-1题的概率(把每个队做对1～m-1题的概率加和，并把各队结果相乘）
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 Source
POJ Monthly,鲁小石
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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
double f[1001][31];
double g[1001][31][31];
int main()
{
int n,t,m;
while(~scanf("%d %d %d",&n,&t,&m))
{
if(n+t+m==0)
break;
for(int i=1; i<=t; i++)
for(int j=1; j<=n; j++)
scanf("%lf",&f[i][j]);
memset(g,0,sizeof(g));
for(int i=1; i<=t; i++)
{
g[i][0][0]=1;
for(int j=1; j<=n; j++)
{
g[i][j][0]=g[i][j-1][0]*(1-f[i][j]);
for(int k=1; k<=j; k++)
{
g[i][j][k] = g[i][j -1][k -1] * (f[i][j])+ g[i][j -1][k] * (1- f[i][j]);
}
}
}
double ans=1;
for(int i=1; i<=t; i++)
ans*=1-g[i][n][0];
//ans=1-ans;
double ans2=1;
for(int i=1; i<=t; i++)
{
double ans1=0;
for(int j=1; j<m; j++)
ans1+=g[i][n][j];
ans2*=ans1;
}
ans-=ans2;
printf("%.3f\n",ans);
}
return 0;
}
poj 2151 Check the difficulty of problems 概率dp
原文：http://blog.csdn.net/wweiainn/article/details/43922585