poj 2249 Binomial Showdown

这是一道简单求C(n,m)的题：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
LL Gcd( LL a, LL b )
{
    return b == 0 ? a : Gcd( b , a%b );    
}
LL C( int n , int k )
{
    LL sum1 = 1,sum2 = 1;
    while( k )
    {
        
         sum1 *= k;
         sum2 *= n; 
         LL t = Gcd( sum1 , sum2 );
         sum1 /= t;
         sum2 /= t;
         n--;
         k--;    
    }    
    return sum2;
}
int main(  )
{
    int n,k;
    while( scanf( "%d %d",&n,&k ),n|k )
    {
        k = min( k , n - k );
        if( k == 0 ) puts( "1" );
        else
        {
           printf( "%I64d\n",C( n , k ) );    
        }
    }
    //system( "pause" );
    return 0;
}