poj 2992 Divisors(求组合数的因子个数)

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http://poj.org/problem?id=2992

该题用到的定理：

.任意一个数n可以写成若干个素数的乘积，即 p1^a1 * p2^a2*......*pn^an，它的的约数的个数为 (a1+1)*(a2+1)*......(an+1).

.对于任意一个素数p, n!中含有p的个数为 (n/p + n/p^2 + n/p^3 + ......).

.c(n,k) = n! / ( k! * (n-k)! ).

#include <stdio.h>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#define LL long long#define _LL __int64const int N = 500;int prime[N];int prime_num;bool vis[N];void Prime(){    prime_num = 0;    memset(vis,true,sizeof(vis));    for(int i = 2; i < 500; i++)    {        if(vis[i])        {            for(int j = i*2; j < 500; j += i)                vis[j] = false;        }    }    for(int i = 2; i <= 431; i++)        if(vis[i])            prime[ prime_num++ ] = i;}//n!中含有素数p的个数int cal(int n, int p){    int ans = 0;    int tmp = p;    while(p <= n)    {        ans += n/p;        p = p*tmp;    }    return ans;}int main(){    int n,k;    Prime();    while(~scanf("%d %d",&n,&k))    {        LL ans = 1;        for(int i = 0; i < prime_num; i++)        {            int a = cal(n,prime[i]);            int b = cal(k,prime[i]);            int c = cal(n-k,prime[i]);            ans *= (a-b-c+1);        }        printf("%lld\n",ans);    }    return 0;}

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