pku2992 Divisors

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

数学，质数，快速求n!中质数p的个数

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

int valid[441] = {0};
int ans[100], tot = 0;

//O(N)筛质数表
void get_prime(int n)
{
    int i, j;
    for(i=2; i<=n; i++)
    {
        if(valid[i] == 0)
        {
            tot ++;
            ans[tot] = i;
        }
        for(j=1; ((j<=tot) && (i*ans[j]<=n)); j++)
        {
            valid[i*ans[j]] = 1;
            if(i%ans[j] == 0)
            {
                break;
            }
        }
    }
}

//求n!中质数p个个数
int f(int n, int p)
{
    return ( n<p? 0: n/p + f(n/p, p) );
}

int main()
{
    long long r;
    int i, x, y;
    get_prime(440);
    while(~scanf("%d%d", &x, &y))
    {
        r = 1;
        for(i=1; i<=tot && ans[i] <= x; i++) 
        {
            r *= (f(x, ans[i]) - f(y, ans[i]) - f(x-y, ans[i]) + 1);
        }
        printf("%lld\n", r);
    }
    return 0;
}