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用整除分塊預處理      

發現對於連續的d , n/d的取值一樣 , 也就是說   的取值一樣 , 然後就可以整除分塊做了

#include<bits/stdc++.h>
#define N 50050
#define LL long long
using namespace std;
int T,n,m,tot; LL s_mu[N],s_f[N];
int mu[N],p[N],isp[N]; 
void Solve(){
	scanf("%d%d",&n,&m); if(n>m) swap(n,m);
	LL ans = 0;
	for(int l=1,r;l<=n;l=r+1){
		int x1=n/l,x2=m/l; r = min(n/x1,m/x2);
		ans += (s_mu[r]-s_mu[l-1]) * s_f[x1] * s_f[x2];
	} printf("%lld\n",ans);
}
int main(){
	mu[1]=s_mu[1]=1;
	for(int i=2;i<=N-50;i++){
		if(!isp[i]) p[++tot]=i,mu[i]=-1;
		for(int j=1;j<=tot;j++){
			if(i*p[j]>N-50) break; isp[i*p[j]]=1; 
			if(i%p[j]==0) break; mu[i*p[j]] = -mu[i];
		}
		s_mu[i] = s_mu[i-1] + mu[i];
	}
	for(int i=1;i<=N-50;i++){ int ans=0;
		for(int l=1,r;l<=i;l=r+1){
			r = i/(i/l);
			ans += (r-l+1) * (i/l);
		} s_f[i] = ans;
	}
	scanf("%d",&T); while(T--) Solve(); 
	return 0;
}