P3327 [sdoi2015] approximate sum (Mobius inversion + formula)

P3327 [SDOI2015] About a number and

The question ：

set up d(x) by x The approximate number of , Given $n,m$, seek
$$\sum _{i=1}^n\sum _{j=1}^{m}d(ij)$$

Solution:

First you need to know a derivation ：
$$d(ij)=\sum _{x|i}\sum _{y|j}[gcd(x,y)=1]$$
And then we start the derivation of the formula ：
$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{m}d(ij) \\ =\sum _{i=1}^n\sum _{j=1}^m\sum _{x|i}\sum _{y|j}[gcd(x,y)=1] \\ =\sum _{x=1}^n\sum _{y=1}^m\lfloor \frac{n}{x}\rfloor \lfloor \frac{x}{m}\rfloor [gcd(x,y)=1] \\ =\sum _{x=1}^n\sum _{y=1}^m\lfloor \frac{n}{x}\rfloor \lfloor \frac{x}{m}\rfloor \sum _{d|gcd(x.y)}\mu (d) \\ =\sum _{d=1}^{min(n,m)}\mu (d)\sum _{x=1}^{\lfloor \frac{n}{d}\rfloor }\lfloor \frac{n}{dx}\rfloor \sum _{y=1}^{\lfloor \frac{m}{d}\rfloor }\lfloor \frac{m}{dy}\rfloor \end{gathered}$$
Here we can see that we can solve it by dividing it into blocks .

Code

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>P;
typedef long long ll;
const int N=5e4;
int mu[N+5],prime[N+5],notPrime[N+5],cnt=0,sumMu[N+5];
ll sum[N+5];
void initMu()
{
    
    mu[1]=1;
    for(int i=2;i<=N;i++)
    {
    
        if(!notPrime[i])
        {
    
            prime[cnt++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<cnt&&1ll*i*prime[j]<=N;j++)
        {
    
            notPrime[i*prime[j]]=1;
            if(i%prime[j])mu[i*prime[j]]=-mu[i];
            else break;
        }
    }
    for(int i=1;i<=N;i++)
    {
    
        sumMu[i]=sumMu[i-1]+mu[i];
        for(int l=1,r;l<=i;l=r+1)
        {
    
            r=i/(i/l);
            sum[i]+=1ll*i/l*(r-l+1);
        }
    }
}
int n,m,t;
int main()
{
    
    initMu();
    scanf("%d",&t);
    while(t--)
    {
    
        scanf("%d%d",&n,&m);
        ll res=0;
        int M=min(n,m);
        for(int l=1,r;l<=M;l=r+1)
        {
    
            r=min(n/(n/l),m/(m/l));
            res+=1ll*(sumMu[r]-sumMu[l-1])*sum[n/l]*sum[m/l];
        }
        printf("%lld\n",res);
    }
    return 0;
}