[Mobius inversion]

A. Luogu P2522 [HAOI2011]Problem b

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=50000+5;
int mu[maxn],prime[maxn],cnt,vis[maxn],n,m,f[maxn],k,a,b,c,d;
void init(){
    // Linear sieve Mobius function 
	f[1]=mu[1]=1;
	for(int i=2;i<=maxn;i++){
    
		if(vis[i]==0){
    
		    prime[++cnt]=i;
		    mu[i]=-1;
		}
		for(int j=1;j<=cnt;j++){
    
			if(i*prime[j]>maxn)break;
			vis[i*prime[j]]=1; 
			mu[i*prime[j]]=i%prime[j]?-mu[i]:0;
			if(i%prime[j]==0)break;
		}
		f[i]=f[i-1]+mu[i];
	}
}
int getsum(int x,int y){
    
	int ans=0;
	for(int l=1,r;l<=min(x,y);l=r+1){
    
		r=min(x/(x/l),y/(y/l));
		ans+=(f[r]-f[l-1])*(x/l)*(y/l);
	}
	return ans;
}
int main(){
    
	init();
	int T;
	scanf("%d",&T);
	while(T--){
    
		scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
	    printf("%d\n",getsum(b/k,d/k)-getsum((a-1)/k,d/k)-getsum((c-1)/k,b/k)+getsum((a-1)/k,(c-1)/k));
    }
}

B. Luogu P3327 [SDOI2015] About a number and （ The number of factors for each number of preprocessing ）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=20101009;
const int maxn=1e7+5;
int mu[maxn],prime[maxn],cnt,vis[maxn],n,m,k,t[maxn],d[maxn];
ll he[maxn],f[maxn];
void init(int k) {
    //mu[]: Mobius ,f[]:mu[] The prefix and ,d[]: Number of factors for each number ,he[]:d[] And ,t[]: The prime factor with the smallest number  
    vis[0]=vis[1]=mu[1]=d[1]=1; 
    for(int i=2;i<=k;i++) {
    
        if (!vis[i])prime[++cnt]=i,mu[i]=-1,d[i]=2,t[i]=1;
		for(int j=1;j<=cnt&&i*prime[j]<=k;j++) {
    
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
    
                mu[i*prime[j]]=0;
				d[i*prime[j]]=d[i]/(t[i]+1)*(t[i]+2);
                t[i*prime[j]]=t[i]+1;
                break;
            }
		    else mu[i*prime[j]]=-mu[i],d[i*prime[j]]=d[i]<<1,t[i*prime[j]]=1;
        }
    }
    for(int i=1;i<=k;i++)f[i]=f[i-1]+mu[i],he[i]=he[i-1]+d[i];
}
int main(){
    
	init(50000);
	int T;
	scanf("%d",&T);
	while(T--){
    
		ll ans=0;
	    scanf("%d%d",&n,&m);
	    if(n>m)swap(n,m);
	    for(int l=1,r;l<=n;l=r+1){
    
		    r=min(n/(n/l),m/(m/l));
		    ans+=1ll*(f[r]-f[l-1])*he[n/l]*he[m/l];
	    }
	    printf("%lld\n",ans);
	}
}