Hillan and the girl —— 莫比乌斯对两个累加的优化

Problem Description
“WTF! While everyone has his girl(gay) friend, I only have my keyboard!” Tired of watching others’ affair, Hillan burst into scream, which made him decide not to hold it back.
“All right, I am giving you a question. If you answer correctly, I will be your girl friend.” After listening to Hillan, Girl replied, “What is the value of ∑ni=1∑mj=1f(i,j), where f(i,j)=0 if gcd(i,j) is a square number and f(i,j)=1 if gcd(i,j) is not a square number(gcd(i,j) means the greatest common divisor of x and y)?”
But Hillan didn’t have enough Intelligence Quotient to give the right answer. So he turn to you for help.

Input
The first line contains an integer T(1≤T≤10,000)——The number of the test cases.
For each test case, the only line contains two integers n,m(1≤n,m≤10,000,000) with a white space separated.

Output
For each test case, the only line contains a integer that is the answer.

Sample Input
2
1 2333333
10 10

Sample Output
0
33

Hint

In the first test case, obviously $f\left(i,j\right)$ always equals to 0, because $i$ always equals to 1 and $\gcd\left(i,j\right)$ is always a square number(always equals to 1).

这个有两个玩意，所以它构造出来就是
$\displaystyle\sum_{sqrt(k)=1}^{n} f(k)=$$\displaystyle\sum_{k|p}^{} g(p)(p<=N)$
然后用莫比乌斯反演就可以得出我们要求的是
$\displaystyle\sum_{k=1}^{sqrt(n)}$$\displaystyle\sum_{k|p}^{} μ(p/k)*f(p)$
而f(p)是在n,m中gcd=p的倍数所有数量，那么就等于
$\displaystyle\sum_{k=1}^{sqrt(n)}$$\displaystyle\sum_{k|p}^{} μ(p/k)*(n/p)*(m/p)$
因为$μ(p/k)$我们是可以预处理出来的，所以需要变换一下式子
$\displaystyle\sum_{p=1}^{n}(n/p)*(m/p)$$\displaystyle\sum_{k|p}^{} μ(p/k)$
n/p和m/p可以通过整数分块加速

#include<bits/stdc++.h>
using namespace std;
#define eps 1e-6
#define ll long long
int n,m;
const int maxn=1e7+5;
int nprime[maxn],prime[maxn],cnt,mu[maxn],sum[maxn];
void init()
{
    cnt=0;
    mu[1]=1;
    for(int i=2;i<maxn;i++)
    {
        if(!nprime[i])
        {
            prime[++cnt]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=cnt&&prime[j]*i<maxn;j++)
        {
            nprime[prime[j]*i]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=sqrt(maxn)+eps;i++)
        for(int j=i*i;j<maxn;j+=i*i)
            sum[j]+=mu[j/i/i];
    for(int i=2;i<maxn;i++)
        sum[i]+=sum[i-1];
}
int main()
{
    int t;
    scanf("%d",&t);
    init();
    while(t--)
    {
        scanf("%d%d",&n,&m);
        ll ans=0;
        int ne;
        if(n>m)
            swap(n,m);
        for(int i=1;i<=n;i=ne+1)
        {
            int l=n/i,r=m/i;
            ne=min(n/l,m/r);
            ans+=(ll)(sum[ne]-sum[i-1])*l*r;
        }
        printf("%lld\n",(ll)n*m-ans);
    }
    return 0;
}