Bin & Jing in wonderland(概率，组合数学)

Problem 2103 Bin & Jing in wonderland

Accept: 201 Submit: 1048 Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Bin has a dream that he and Jing are both in a wonderland full of beautiful gifts. Bin wants to choose some gifts for Jing to get in her good graces.

There are N different gifts in the wonderland, with ID from 1 to N, and all kinds of these gifts have infinite duplicates. Each time, Bin shouts loudly, “I love Jing”, and then the wonderland random drop a gift in front of Bin. The dropping probability for gift i (1≤i≤N) is P(i). Of cause, P(1)+P(2)+…+P(N)=1. Bin finds that the gifts with the higher ID are better. Bin shouts k times and selects r best gifts finally.

That is, firstly Bin gets k gifts, then sorts all these gifts according to their ID, and picks up the largest r gifts at last. Now, if given the final list of the r largest gifts, can you help Bin find out the probability of the list?

Input

The first line of the input contains an integer T (T≤2,000), indicating number of test cases.

For each test cast, the first line contains 3 integers N, k and r (1≤N≤20, 1≤k≤52, 1≤r≤min(k,25)) as the description above. In the second line, there are N positive float numbers indicates the probability of each gift. There are at most 3 digits after the decimal point. The third line has r integers ranging from 1 to N indicates the finally list of the r best gifts’ ID.

Output

For each case, output a float number with 6 digits after the decimal points, which indicates the probability of the final list.

Sample Input

4 2 3 3 0.3 0.7 1 1 1 2 3 3 0.3 0.7 1 1 2 2 3 3 0.3 0.7 1 2 2 2 3 3 0.3 0.7 2 2 2

Sample Output

0.027000 0.189000 0.441000 0.343000

Source

“高教社杯”第三届福建省大学生程序设计竞赛

题解：有N种苹果，一个人在下面喊k次，每次掉下来一个苹果，每种苹果掉下来的概率已知，现在给出前r大的苹果种类；

问此时的概率；先找出最小的出现的种类mi，求出比mi大的概率，再求出比mi小的概率，枚举mi出现的次数，概率想乘就好了；

代码：

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

using namespace std;

double p[], dp[];

typedef long long LL;

LL C[][];

int num[];

int a[];

void db(){

    C[][] = C[][] = ;

    for(int i = ; i < ; i++){

        C[i][] = C[i][i] = ;

        for(int j = ; j < i; j++){

            C[i][j] = C[i - ][j] + C[i - ][j - ];

        }

    }

}

int main(){

    int T, N, k, r;

    scanf("%d", &T);

    db();

    while(T--){

        scanf("%d%d%d", &N, &k, &r);

        dp[] = ;

        for(int i = ; i <= N; i++){

            scanf("%lf", p + i);

            dp[i] = dp[i - ] + p[i];

        }

        double ans = ;

        memset(num, , sizeof(num));

        int mi = 0x3f3f3f3f;

        for(int i = ; i <= r; i++){

            scanf("%d", a + i);

            num[a[i]]++;

            mi = min(mi, a[i]);

        }

        int now = k;

        for(int i = mi + ; i <= N; i++){

            if(!num[i])continue;

            ans *= C[now][num[i]] * pow(p[i], num[i]);

            now -= num[i];

        }

        double ans1 = ;

        for(int i = num[mi]; i <= k - r + num[mi]; i++){

            ans1 += C[k - r + num[mi]][i] * pow(p[mi], i) * pow(dp[mi - ], k - r + num[mi] - i);

        }

        printf("%.6lf\n", ans * ans1);

    }

    return ;

}