P3768 简单的数学题（莫比乌斯反演）

[题目链接] https://www.luogu.org/problemnew/show/P3768

[题目描述]

求

\(\sum_{i=1}^{n}\sum_{j=1}^{n}i* j* gcd(i,j)\mod\ p\)

[欧拉反演题解] https://www.luogu.org/blog/zhoutb2333/solution-p3768

/*
-----------------------
最大测试点,时限6s
[Input]
1000000007 9786510294
[Output]
27067954
-----------------------2019.2.24
*/
#include<bits/stdc++.h>
#include<tr1/unordered_map>
//#define int long long
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
	register LL x=0,f=1;register char c=getchar();
	while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
	while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
	return f*x;
}
const int MAXN=5e6+5;
int phi[MAXN],prime[MAXN];bool vis[MAXN];
LL sphi[MAXN];
LL n,mod,inv6,ans;
unordered_map <LL,LL> Sphi;
inline LL qpow(LL a,LL b){
	LL res=1;
	while(b){
		if(b&1) (res*=a)%=mod;
		(a*=a)%=mod;
		b>>=1;
	}
	return res;
}
inline void init(int n){
	phi[1]=1;
	for(int i=2;i<=n;i++){
		if(!vis[i]){
			prime[++prime[0]]=i;
			phi[i]=i-1;
		}
		int x;
		for(int j=1;j<=prime[0]&&(x=i*prime[j])<=n;j++){
			vis[x]=1;
			if(i%prime[j]==0){
				phi[x]=phi[i]*prime[j];
				break;
			}
			phi[x]=phi[i]*phi[prime[j]];
		}
	}
	for(int i=1;i<=n;i++){
		sphi[i]=(sphi[i-1]+1ll*phi[i]*i%mod*i%mod)%mod;
	}
}
inline LL s2(LL x){
	x%=mod;
	return x*(x+1)%mod*(2*x+1)%mod*inv6%mod;
}
inline LL s3(LL x){
	x%=mod;
	return ((x*(x+1)/2)%mod)*((x*(x+1)/2)%mod)%mod;
}
inline LL S_phi(LL n){
	if(n<MAXN) return sphi[n];
	if(Sphi[n]) return Sphi[n];
	LL res=s3(n);
	for(LL l=2,r;l<=n;l=r+1){
		r=n/(n/l);
		res=(res-(1ll*(s2(r)-s2(l-1)+mod)%mod*S_phi(n/l)%mod)+mod)%mod;
	}
	return Sphi[n]=res;
}
int main(){
	mod=read(),n=read();
	inv6=qpow(6,mod-2);
	init(MAXN-1);
	for(LL l=1,r;l<=n;l=r+1){
		r=n/(n/l);
		(ans+=1ll*(S_phi(r)-S_phi(l-1)+mod)%mod*s3(n/l)%mod)%=mod;
	}
	printf("%lld\n",ans);
}