个人模板 组合数 C（m,n）

#include <bits/stdc++.h>  

const int maxn=1000000; 
#include <vector>  
using namespace std; 
bool arr[maxn+1]={false}; 
vector<int> produce_prim_number() 
{ 
        vector<int> prim; 
        prim.push_back(2); 
        int i,j; 
        for(i=3;i*i<=maxn;i+=2) 
        { 
                if(!arr[i]) 
                { 
                        prim.push_back(i); 
                        for(j=i*i;j<=maxn;j+=i) 
                                arr[j]=true; 
                } 
        } 
        while(i<maxn) 
        { 
                if(!arr[i]) 
                        prim.push_back(i); 
                i+=2; 
        } 
        return prim; 
} 
//计算n!中素数因子p的指数  
int cal(int x,int p) 
{ 
        int ans=0; 
        long long rec=p; 
        while(x>=rec) 
        { 
                ans+=x/rec; 
                rec*=p; 
        } 
        return ans; 
} 
//计算n的k次方对m取模，二分法  
int pow(long long n,int k,int M) 
{ 
        long long ans=1; 
        while(k) 
        { 
                if(k&1) 
                { 
                        ans=(ans*n)%M; 
                } 
                n=(n*n)%M; 
                k>>=1; 
        } 
        return ans; 
} 
//计算C（n，m）  
int combination(int n,int m) 
{ 
        const int M=10007; 
        vector<int> prim=produce_prim_number(); 
        long long ans=1; 
        int num; 
        for(int i=0;i<prim.size()&&prim[i]<=n;++i) 
        { 
                num=cal(n,prim[i])-cal(m,prim[i])-cal(n-m,prim[i]); 
                ans=(ans*pow(prim[i],num,M))%M; 
        } 
        return ans; 
} 
int main() 
{ 
        int m,n; 
        while(~scanf("%d%d",&m,&n),m&&n) 
        { 
                printf("%d\n",combination(m,n)); 
        } 
        return 0; 
} 

#include<bits/stdc++.h>
using namespace std;

const int N = 10000 + 10;
int n, m;

int xx(int m)   /// O(n)计算C(m,n)
{
    int c[N];
    c[0] = 1;
    for(int i = 1; i <= m; i++)
    {
        c[i] = c[i-1] * (n-i+1)/i;
    }
    return c[m];
}

int main()
{
    while(scanf("%d%d", &n, &m) == 2)
    {
        printf("%d\n", xx(m));
    }
}