hdu 5894(组合数取模)

Posted 2020-08-09 AC菜鸟机

hannnnah_j’s Biological Test

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 681    Accepted Submission(s): 235

Problem Description

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

 

 

Source

2016 ACM/ICPC Asia Regional Shenyang Online

 

题意:n个位子,m个人,每两个人之间至少隔k个位置,问排的方式?

题解:设第一个人和第二个人之间的距离为 a1,第二个和第三个之间距离为 a2 ... 第 n个人和第1 个人之间距离为 an;

a1+...+an = n-m....1

ai>=k                ....2

解方程 1,2得解为 C(n-m*k-1,m-1)

第一个人有n种放法,所以答案要乘n,m个人没有区别,答案要除 m.

这场比赛真心不爽,1010坑了3个小时,然后这个题目最后把组合数写出来了,刚好59分快5点了,然后没交上去....交上去...事后在hdu AC....又是2个题..

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const long long  p =  1e9+7;
typedef long long LL;
LL pow_mod(LL a,LL n)
{
    LL ans=1;
    while(n)
    {
        if(n&1) ans=ans*a%p;
        a=a*a%p;
        n>>=1;
    }
    return ans;
}
LL cm(LL n,LL m,LL mod)
{
    if(m>n) return 0;
    LL i,ans=1,a,b;
    for(i=0; i<m; i++)
    {
        a=(n-i)%mod;
        b=(m-i)%mod;
        ans=ans*( a*pow_mod(b,mod-2)%mod )%mod;
    }
    return ans;
}
LL lucas(LL n,LL m,LL p)
{
    if(m==0) return 1;
    return ( cm(n%p,m%p,p)*lucas(n/p,m/p,p) )%p;
}

int main()
{
    int T;
    long long n,m,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld %lld %lld",&n,&m,&k);
        LL a = lucas(n-m*k-1,m-1,p);
        LL c = pow_mod(m,p-2)%p;
        printf("%lld\n",((a*n)%p*c)%p);

    }
    return 0;
}