hannnnah_j’s Biological Test HDU - 5894 （组合数）-优快云博客

本文链接：https://blog.youkuaiyun.com/Coldfresh/article/details/77893247

本文介绍了一个组合数学问题，即如何计算在圆形桌子周围安排学生就座的方法数，要求每两个学生之间至少有k个空位。文章提供了一段C++代码实现，通过数学计算解决了这一问题。

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 hannnnah_j is a teacher in WL High school who teaches biology.

One day, she wants to test m students, thus she arranges n different seats around a round table.

In order to prevent cheating, she thinks that there should be at least k empty seats between every two students.

hannnnah_j is poor at math, and she wants to know the sum of the solutions.So she turns to you for help.Can you help her? The answer maybe large, and you need to mod 1e9+7.

Input
First line is an integer T(T≤1000).
The next T lines were given n, m, k, respectively.
0 < m < n < 1e6, 0 < k < 1000
Output
For each test case the output is only one integer number ans in a line.
Sample Input

2
4 2 6
5 2 1

Sample Output

0
5

这题是一个组合数学的题，因为每个椅子是不一样的，所以其实看成链状的而不是环，我们假定链的第一个位置已经坐了人，并且顺带带走了k*m个点，而且我们把1+k个点浓缩成一个点，那么m-1个这样的点来放剩下的n-1-k*m个格子即可，然后我们可以截成n种不同的链所以要乘n，但是有因为是循环放置的，而且每个人是相同的所以通构的会有m种，因为是旋转的，所以要除m。
m等于1的时候要特判

#include<iostream>
#include<cstdio>
#include<cstring>
#define mod 1000000007
#define N 1000005
using namespace std;
long long fac[N];
long long inv[N];
long long P(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b&1)
            ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
long long get(int n,int m)
{
    return (fac[n]*inv[m]%mod)*inv[n-m]%mod;
}
void init()
{
    fac[0]=1;
    inv[0]=1;
    for(int i=1;i<N;i++)
        fac[i]=fac[i-1]*i%mod;
    inv[N-1]=P(fac[N-1],mod-2);
    for(int i=N-2;i>=1;i--)
        inv[i]=inv[i+1]*(i+1)%mod;
}

int main()
{   
    init();
    int n,m,k;
    int t;
    //cout<<get(10,5)<<endl;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        if(m==1)
            printf("%d\n",n);
        else
            if(n<(k+1)*m)
                printf("0\n");
            else
            {
                long long ans=get(n-1-k*m,m-1);
                ans=ans*n%mod;
                ans=ans*P(m,mod-2)%mod;
                printf("%lld\n",ans);
            }   
    }   
    return 0;
}