LightOj 1341

題目大意：
s=a∗b ; 已知s,amin (最小的a )問有多少個這樣的組合；a≠b (沒看見錯了好幾次);
思路：
s=∏i=1kpeii;s 質數分解;
Ns的因子個數 ,=∑i=1k(1+ei) ;
num(a,b)=N2 ;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define N 1000010
#define LL long long

using namespace std;

bool not_prime[N];
int prim[100000];
int cnt = 0;

void init()
{
    memset(not_prime, 0, sizeof(not_prime));

    prim[1] = 0;

    for(int i = 2; i < N; i++) 
    {  
        if(!not_prime[i]) 
        {   
            prim[cnt++] = i;
            for(int j = i + i; j < N; j += i)
            {  
                not_prime[j] = 1;  
            }  
        }  
    } 

    // cout << cnt << ' ' << prim[cnt-1] << endl; 
}

LL get(LL n)
{
    LL ans = 1;

    for (int i = 0; prim[i] * prim[i] <= n && i < cnt; i++)
    {
        if (n % prim[i] == 0)
        {
            int cnt = 0;

            while (n % prim[i] == 0)
            {
                cnt++;
                n /= prim[i];
            }

            ans *= (cnt + 1);
        }
    }

    if (n > 1)
    {
        ans *= 2;
    }

    return ans;
}

int main()
{
    init();

    int T;
    scanf("%d", &T);

    for (int cas  = 1; cas <= T; cas++)
    {
        LL n, m;
        scanf("%lld%lld", &n, &m);

        if (n <= m * m)
        {
            printf("Case %d: 0\n", cas);
            continue;
        }

        LL ans = get(n) / 2;

        for (int i = 1; i < m; i++)
        {
            if (n % i == 0)
            {
                ans--;
            }
        }

        printf("Case %d: %lld\n", cas, ans);
    }
    return 0;
}