3985. I guess the gift is a bag!

原创于 2016-07-11 00:27:52 发布 · 277 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#TJU

数学问题专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一个具体的概率问题：如何计算在有限数量的背包中，特定的朋友能够每人获得一个背包的概率，并提供了一种通过组合数学公式解决该问题的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Boss Yu said that he will give gifts to the members who passed this TJU ACM recruitment contest and I guess the gifts will be some bags.
But here comes the problem: suppose there are N people and M bags. I have K good friends among the N people. Now I want to know what’s the possibility of each of the K people can have a bag.
Can you tell me the possibility?
Input

The first line contains a single integer T represents the number of test cases, for each test case there are only three integers N, M and K(1≤K≤M≤N≤1000,T≤100000).
Output

The possibility of each of the K people can have a bag, rounded to four decimal places.
Sample Input

3
2 1 1
21 3 2
1000 500 2
Sample Output

0.5000
0.0143
0.2497

题意：给出总人数，和总的背包数，以及朋友数目，求每一个朋友都能被分到一个背包的概率。

直接做显然不行,然后就推公式啦。左边是朋友都有书包的概率。

C m - k n - k C m n = a n s * n - i + 1 i

${C^{m-k}_{n-k}\over{C^m_n}}=ans*{{n-i+1}\over{i}}$

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
double c(int n,int k)
{
    double ans=1;
    for(int i=1;i<=k;i++)
    {
        ans*=n-i+1;
        ans/=i;
    }
    return ans; 
}

int main (void)
{
    int t;
    cin>>t;
    while(t--)
    {
        int m,n,k;
        scanf("%d %d %d",&n,&m,&k);
        printf("%.4f\n",c(n-k,m-k)/c(n,m));
    }
    return 0;
}