大组合数取模

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1119

考虑从(1,1)->(n,m)必定会向下走n-1步,向右走m-1步,那么总的走法是C(n-1+m-1,m-1)。

关于组合数取模:大神博客：http://blog.csdn.net/acdreamers/article/details/8037918

#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;
typedef long long LL;
const LL p = 1e9+7;
LL n,m;

LL quick_mod(LL a, LL b)
{
    LL ans = 1;
    a %= p;
    while(b)
    {
        if(b & 1)
        {
            ans = ans * a % p;
            b--;
        }
        b >>= 1;
        a = a * a % p;
    }
    return ans;
}

LL C(LL n, LL m)
{
    if(m > n) return 0;
    LL ans = 1;
    for(int i=1; i<=m; i++)
    {
        LL a = (n + i - m) % p;
        LL b = i % p;
        ans = ans * (a * quick_mod(b, p-2) % p) % p;
    }
    return ans;
}

LL Lucas(LL n, LL m)
{
    if(m == 0) return 1;
    //printf("%I64d
",C(n,m));
    return C(n % p, m % p) * Lucas(n / p, m / p) % p;
}

int main()
{
    scanf("%I64d%I64d", &n, &m);
    printf("%I64d
", Lucas(n+m-2,m-1));
    return 0;
}