SPOJ PGCD - Primes in GCD Table (莫比烏斯)

//題意：給定兩個數和，其中，，求爲質數的有多少對？其中和的範

     圍是。

//方法：莫比烏斯反演

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn = 1e7 + 5;
int isprime[maxn];
int p[maxn];
int u[maxn];
int g[maxn];
int sum[maxn];
int cnt;
void mobi(int n)
{
    memset(isprime, 0, sizeof isprime);
    
    u[1] = 1;
    cnt = 0;
    for(int i = 2; i <= n; i++)
    {
        if(!isprime[i])
        {
            p[cnt++] = i;
            u[i] = -1;
            g[i] = 1;
        }
        for(int j = 0; j < cnt && i * p[j] <= n; j++)
        {
            isprime[i * p[j]] = 1;
            if(i % p[j])
            {
                u[i * p[j]] = -u[i];
                g[i * p[j]] = u[i] - g[i];
            }
            else{
                u[i * p[j]] = 0;
                g[i * p[j]] = u[i];
                break;
            }
        }
    }
    sum[0]= 0;
    for(int i = 1; i <= n; i++)
        sum[i] = sum[i - 1] + g[i];
}
int main()
{
    int t;
    cin>>t;
    mobi(10000000);
    while(t--)
    {
        int n, m;
        cin>>n>>m;
        if(n > m)
            swap(n, m);
        long long ans = 0;
        for(int i = 1, last; i <= n; i = last + 1)
        {
            last = min(n/(n/i), m/(m/i));
            ans += 1ll * (n / i) * (m / i) * (sum[last] - sum[i -1]);
        }
        cout<<ans<<endl;
    }
    return 0;
}