3985. I guess the gift is a bag!

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。

Cm−kn−kCmn=ans∗n−i+1i

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;}