[BZOJ2226]LCMSum

转化一下，$\sum\limits_{i=1}^n[i,n]=n\sum\limits_{i=1}^n\dfrac i{(i,n)}$

枚举$d=(i,n)$，上式变为$n\sum\limits_{d=1}^n\sum\limits_{i=1}^n[(i,n)=d]\dfrac id=n\sum\limits_{d|n}\sum\limits_{i=1}^{\frac nd}\left[\left(i,\dfrac nd\right)=1\right]i$

设$f(n)=\sum\limits_{i=1}^n[(i,n)=1]i$，即互质数和

$$\begin{align*}f(n)&=\sum\limits_{i=1}^ni\sum\limits_{d|(i,n)}\mu(d)\\&=\sum\limits_{d|n}\mu(d)\sum\limits_{\substack{d|i\\i\leq n}}i\\&=\sum\limits_{d|n}d\mu(d)\sum\limits_{i=1}^{\frac nd}i\\&=\dfrac n2\sum\limits_{d|n}\mu(d)\left(\dfrac nd+1\right)\\&=\dfrac n2\left([n=1]+\sum\limits_{d|n}\mu(d)\dfrac nd\right)\\&=\dfrac n2\left([n=1]+\varphi(n)\right)\end{align*}$$

最后一步转变的依据可以用$n=\sum\limits_{d|n}\varphi(d)$反演得到

于是我们可以$O(1)$算$f(n)$了，原式变成$n\sum\limits_{d|n}f\left(\dfrac nd\right)=n\sum\limits_{d|n}f(d)$，$O(\sqrt n)$枚举约数就好了

#include<stdio.h>
#define ll long long
#define T 1000000
int phi[1000010],pr[1000010];
bool np[1000010];
void sieve(){
	int i,j,m=0;
	np[1]=1;
	phi[1]=1;
	for(i=2;i<=T;i++){
		if(!np[i]){
			m++;
			pr[m]=i;
			phi[i]=i-1;
		}
		for(j=1;j<=m;j++){
			if(pr[j]*(ll)i>T)break;
			np[i*pr[j]]=1;
			if(i%pr[j]==0){
				phi[i*pr[j]]=phi[i]*pr[j];
				break;
			}else
				phi[i*pr[j]]=phi[i]*(pr[j]-1);
		}
	}
}
ll f(int n){return(phi[n]+(n==1))*(ll)n/2;}
int main(){
	sieve();
	int t,i,n;
	ll s;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		s=0;
		for(i=1;i*i<=n;i++){
			if(n%i==0){
				s+=f(n/i);
				if(i*i<n)s+=f(i);
			}
		}
		printf("%lld\n",n*s);
	}
}