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:一套题,有T个题,M个团队参加比赛,已知每个团队做出来某题的概率。问你,你能计算出所有团队至少解决一个问题,同时冠军团队解决至少N问题的概率吗?
思路:可以把所有考生都至少做来一道题的概率减去 每个人都做来1到n-1道题的概率。
p=[(1-x11)*(1-x12)(..) ] * [(1-x21)*(1-x22)(..)]*[...] - [...]*[...] ,这样的话,用组合数就ok了。
但是这里是用的DP是思路,先把考生与考题的关系求出来,dp[i][j][k] 表示第i个考试前j个题会做k道的概率。再根据题意进行分情况DP。
AC代码:
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
#include<iomanip>
const int maxx=1010;
const int inf=0x3f3f3f3f;
using namespace std;
double dp[maxx][35][35];
double a[maxx][35];
int main()
{
int m,t,n;
while(cin>>m>>t>>n)
{
if(m==0 && t==0 && n==0)
break;
memset(dp,0,sizeof(dp));
for(int i=1; i<=t; i++)
{
for(int j=1; j<=m; j++)
{
cin>>a[i][j];
}
}
for(int i=1; i<=t; i++)
{
for(int j=0; j<=m; j++)
{
for(int k=0; k<=j; k++)
{
if(j==0 && k==0)
dp[i][j][k]=1;
else if(j==k)
dp[i][j][k]=dp[i][j-1][k-1]*a[i][j];
else if(k==0)
dp[i][j][k]=dp[i][j-1][k]*(1-a[i][j]);
else
dp[i][j][k]=dp[i][j-1][k-1]*a[i][j]+dp[i][j-1][k]*(1-a[i][j]);
}
}
}
double ans=1,p1=1,tmp;
for(int i=1; i<=t; i++)
{
p1*=(1-dp[i][m][0]);
tmp=0;
for(int j=1; j<n; j++)
tmp+=dp[i][m][j];
ans*=tmp;
}
ans=p1-ans;
cout<<fixed<<setprecision(3)<<ans<<endl;
}
return 0;
}