# POJ2151 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 $P_{ij}(1 \leqslant i \leqslant T, 1\leqslant j \leqslant M)$. 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 $M (0 < M \leqslant 30), T (1 < T \leqslant 1000)$ and $N (0 < N \leqslant M)$. 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 $P_{ij}$. 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


/*program from Wolfycz*/
#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define lMax 1e18
#define MK make_pair
#define iMax 0x7f7f7f7f
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0)	putchar('-'),x=-x;
if (x>9)	print(x/10);
putchar(x%10+'0');
}
const int N=1e3,M=30;
double P[N+10][M+10],F[N+10][M+10];
int main(){
while (true){
//m:problems n:teams
memset(F,0,sizeof(F));
if (!(n+m+k))   break;
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
scanf("%lf",&P[i][j]);
for (int i=1;i<=n;i++){
F[i][0]=1;
for (int j=1;j<=m;j++){
for (int k=j;k;k--)
F[i][k]=F[i][k-1]*P[i][j]+F[i][k]*(1-P[i][j]);
F[i][0]*=(1-P[i][j]);
}
for (int j=1;j<=m;j++)  F[i][j]+=F[i][j-1];
}
double Ans1=1,Ans2=1;
for (int i=1;i<=n;i++)  Ans1*=1-F[i][0];
for (int i=1;i<=n;i++)  Ans2*=F[i][k-1]-F[i][0];
printf("%.3f\n",Ans1-Ans2);
}
return 0;
}

posted @ 2022-01-12 21:55  Wolfycz  阅读(13)  评论(0编辑  收藏  举报