排列组合——lusca定理

http://www.lydsy.com/JudgeOnline/problem.php?id=2982

T行，每行一个数，为C(n, m) mod 10007的答案。(1<=m<=n<=200,000,000)

/**************************************************************
    Problem: 2982
    User: huhuuu
    Language: C++
    Result: Accepted
    Time:236 ms
    Memory:1272 kb
****************************************************************/
 
#include<iostream>
#include<stdio.h>
using namespace std;
 
int pow_mod(int a,int n,int p)
{
    int ans=1,t=a;
    while(n)
    {
        if(n&1)
            ans=(long long)ans*t%p;
        t=(long long)t*t%p;
        n>>=1;
    }
    return ans;
}
int cal(int n,int m,long p)
{
    if(m>n-m) m=n-m;
    int ans=1;
    for(int i=1;i<=m;i++)
    {
        ans=(long long)ans*(n-i+1)%p;
        int a=pow_mod(i,p-2,p);
        ans=(long long)ans*a%p;
    }
    return ans;
}
int com_mod(int n,int m,long p)
{
    if(n<m)return 0;
    return cal(n,m,p);
}
int lucas(int n,int m,long p)
{
    int r=1;
    while(n&&m&&r)
    {
        r=(long long)r*com_mod(n%p,m%p,p)%p;
        n/=p;
        m/=p;
    }
    return r;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,p;
        scanf("%d%d",&n,&m);p=10007;
        printf("%d\n",lucas(n,m,p));
 
    }
}