BZOJ 1040 最大公约数之和

Description

求(sum_{i=1}^n(i,n),nleqslant 10^9)

Solution

(sum_{i=1}^n(i,n)=sum_{dmid n}dsum_{i=1}^n[(i,n)=d]=sum_{dmid n}sum_{i=1}^{frac{n}{d}}[(i,frac{n}{d})=1]=sum_{dmid n}varphi(frac{n}{d}))

复杂度好像和杜教筛挺像的?口胡+1...我不太会算...

Code

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll phi(ll x) {
	ll r=x;
	for(int i=2;i*i<=x && x>1;i++) if(x%i==0) {
		r=r/i*(i-1),x/=i;
		for(;x%i==0;x/=i);
	}if(x>1) r=r/x*(x-1);return r;
}
ll S(ll n) { return n*(n+1)/2; }
ll S(ll l,ll r) { return S(r)-S(l-1); }
ll F(ll n) {
	ll r=0;
	for(int i=1;i*i<=n;i++) if(n%i==0) {
		r=r+i*phi(n/i);
		if(i*i==n) continue;
		r=r+n/i*phi(i);
	}return r;
}
int main() {
	ll n;
	scanf("%lld",&n);
	printf("%lld
",F(n));
	return 0;
}