POJ 2992 Divisors 求一個數的因數的個數

題意：輸入C(n,k), 求該數的因數個數。

題解：對於任意質數p, n!中有（n/p+n/p^2+n/p^3+...)個質因子p。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAX = 450;int a[MAX], p[MAX], pn;void prime (){    pn = 0;    memset(a,0,sizeof(a));    int i, j;    for ( i = 2; i < MAX; i++ )    {        if ( a[i] == 0 )            p[pn++] = i;        for ( j = 0; j < pn && i * p[j] < MAX && (p[j] <= a[i] || a[i] == 0); j++ )            a[i*p[j]] = p[j];    }}int count ( int x, int pri ){    int ret = 0, tmp = pri;    while ( x >= tmp )    {        ret += x / tmp;        tmp *= pri;    }    return ret;}int main(){    int n, k;    prime();    while ( scanf("%d%d",&n,&k) != EOF )    {        int a, b, c;        __int64 res = 1;        for ( int i = 0; i < pn && p[i] <= n; i++ )        {            a = count ( n, p[i] );            b = count ( k, p[i] );            c = count ( n-k, p[i] );            res *= ( a - b - c + 1 );        }        printf("%I64d\n", res );    }    return 0;}