P2257 YY的GCD

求(ans=sumlimits_{i=1}^nsumlimits_{i=1}^m[gcd(i,j)=prime])

设函数(f(d)=sumlimits_{i=1}^nsumlimits_{i=1}^m[gcd(i,j)=d])

设函数$F(d)=sumlimits_{i=1}^nsumlimits_{i=1}^m[d|gcd(i,j)]=leftlfloor frac{n}{d} ight floor leftlfloor frac{m}{d} ight floor $

莫比乌斯反演：

(F(d)=sumlimits_{d|k}f(k))

( herefore f(d)=sumlimits_{d|k}mu(leftlfloor frac{k}{n} ight floor)F(d))

推式子了

(ans=sumlimits_{pin prime}sumlimits_{i=1}^nsumlimits_{j=1}^m[gcd(i,j)=p])

将(f(d)=sumlimits_{i=1}^nsumlimits_{i=1}^m[gcd(i,j)=d])带进去

(ans=sumlimits_{pin prime}f(p))

将(f(d)=sumlimits_{d|k}mu(leftlfloor frac{k}{n} ight floor)F(d))带进去

(ans=sumlimits_{pin prime}sumlimits_{p|k}mu(leftlfloor frac{k}{n} ight floor)F(p))

不觉得(p|k)看着不舒服吗，也不好处理，我们换一种方式枚举(leftlfloor frac{k}{n} ight floor)

(ans=sumlimits_{p in prime} sumlimits_{d=1}^{ min( dfrac{n}{p},frac{m}{p} ) } mu(d) F(dp))

(~~~~~=sumlimits_{p in prime} sumlimits_{d=1}^{ min( frac{n}{p},frac{m}{p} ) } mu(d) leftlfloor frac{n}{dp} ight floor leftlfloor frac{m}{dp} ight floor)

这样做也不好分块，前面的(p)有点难处理，我们应该把素数想办法和莫比乌斯函数联系起来，然后丢到欧拉筛处理

设(T)为(dp)放到前面去

(ans=sumlimits_{T=1}^{min(n,m)} sumlimits_{t|T,t in prime} mu(leftlfloor frac{T}{t} ight floor) leftlfloor frac{n}{T} ight floor leftlfloor frac{m}{T} ight floor)

还是不好分块吧，(mu)单独就好处理了，于是我们变成这样

$ans=sumlimits_{T=1}^{min(n,m)} leftlfloor frac{n}{T} ight floor leftlfloor frac{m}{T} ight floor sumlimits_{t|T,t in prime} mu(leftlfloor frac{T}{t} ight floor) $

后面的(sumlimits_{t|T,t in prime} mu(leftlfloor frac{T}{t} ight floor))预处理出来，然后对前面的(leftlfloor frac{n}{T} ight floor leftlfloor frac{m}{T} ight floor)进行分块就好了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=1e7+10;
inline int Read(){
	LL x=0,f=1; char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-') f=-1; c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<3)+(x<<1)+c-'0',c=getchar();
	}
	return x*f;
}
int T;
int mu[maxn],prime[maxn];
LL g[maxn],sum[maxn];
bool visit[maxn];
inline void F_phi(LL n){
	mu[1]=1;
	int tot=0;
	for(int i=2;i<=n;++i){
		if(!visit[i]){
			prime[++tot]=i;
			mu[i]=-1;
		}
		for(int j=1;j<=tot&&i*prime[j]<=n;++j){
			visit[i*prime[j]]=true;
			if(i%prime[j]==0)
				break;
			else
			    mu[i*prime[j]]=-mu[i];
		}
	}
	for(int j=1;j<=tot;++j)
	    for(LL i=1;i*prime[j]<=n;++i)
	        g[i*prime[j]]+=(LL)mu[i];
	for(int i=1;i<=n;++i)
	    sum[i]=sum[i-1]+g[i];
}
int main(){
	T=Read();
	F_phi(maxn);
	while(T--){
		int n=Read(),m=Read();
		if(n>m)
		    swap(n,m);
		LL ans(0);
		for(int l=1,r;l<=n;l=r+1){
			r=min(n/(n/l),m/(m/l));
			ans+=(LL)(n/l)*(m/l)*(sum[r]-sum[l-1]);
		}
		printf("%lld
",ans);
	}
}