luoguP2257 YY的GCD

题意

设(f(d)=sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=d],F(d)=sumlimits_{k|d}f(k)=frac{n}{d}*frac{m}{d})

得：
(f(n)=sumlimits_{n|d}mu(frac{d}{n})F(d))

(ans=sumlimits_{pin prime}f(p))
(=sumlimits_{pin prime}sumlimits_{p|d}mu(frac{d}{p})*F(d))
设(frac{d}{p}=t)
(=sumlimits_{pin prime}sumlimits_{t=1}^{min(frac{n}{p},frac{m}{p})}mu(t)*F(t*p))
(=sumlimits_{pin prime}sumlimits_{t=1}^{min(frac{n}{p},frac{m}{p})}mu(t)*frac{n}{t*p}*frac{m}{t*p})
设(t*p=T)
(=sumlimits_{T=1}^{min(n,m)}sumlimits_{t|T,tin prime}mu(frac{T}{t})*frac{n}{T}*frac{m}{T})
(=sumlimits_{T=1}^{min(n,m)}frac{n}{T}*frac{m}{T}*(sumlimits_{t|T,tin prime}mu(frac{T}{t})))

于是就可以做了。

code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=10000010;
int T,n,m;
int mu[maxn],sum[maxn];
bool vis[maxn];
vector<int>prime;
inline void shai(int n)
{
	vis[1]=1;mu[1]=1;
	for(int i=2;i<=n;i++)
	{
		if(!vis[i])prime.push_back(i),mu[i]=-1;
		for(unsigned int j=0;j<prime.size()&&i*prime[j]<=n;j++)
		{
			vis[i*prime[j]]=1;
			if(i%prime[j]==0){mu[i*prime[j]]=0;break;}
			mu[i*prime[j]]=-mu[i];
		}
	}
	for(unsigned int i=0;i<prime.size();i++)
		for(int j=1;j*prime[i]<=n;j++)
			sum[j*prime[i]]+=mu[j];
	for(int i=1;i<=n;i++)sum[i]+=sum[i-1];
}
inline ll solve(int n,int m)
{
	ll res=0;
	for(int l=1,r;l<=min(n,m);l=r+1)
	{
		r=min(n/(n/l),m/(m/l));
		res+=1ll*(n/l)*(m/l)*(sum[r]-sum[l-1]);
	}
	return res;
}
int main()
{
	shai(10000000);
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		printf("%lld
",solve(n,m));
	}
	return 0;
}