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

题目大意,给T个队伍,M个题目,给出每个队伍做出每个题的概率\(P_{ij}\),求每个队伍至少做出一道题,且冠军队伍至少做出\(N\)道题的概率

由于每个队伍做题的概率是独立的,所以我们可以提前统计出第\(i\)个队伍至少做出\(k\)个题的概率\(F[i][k]\)

然后我们统计出每个队伍至少做出1道题的概率\(Ans1\),再统计出每个队伍做出\(1~N-1\)的概率\(Ans2\),答案即为\(Ans1-Ans2\)

/*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;
template<typename T>inline T read(T x){
    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){
        int m=read(0),n=read(0),k=read(0);
        //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