求\(\sum_{i=1}^{n}\gcd(i,n)~(1\leq n\leq 10^9)\)
显然有
\[\sum_{i=1}^{n}\gcd(i,n)=\sum_{d|n}d\varphi(\frac nd) \]
(考虑 \(\varphi\) 的意义)
然后直接做随便做。
Code
#include<iostream>
#include<cstdio>
using namespace std;
long long getphi(long long x){
long long ret=1;
for(int i=2;1ll * i*i<=x;++i){
if(x%i==0){
for(ret*=i-1;(x/=i)%i==0;ret*=i);
}
}
return x>1?ret*(x-1):ret;
}
int main(){
int n;
while(scanf("%d",&n) != EOF){
long long ans = 0;
for(int i = 1; 1ll * i * i <= n; ++ i){
if(n % i != 0) continue;
long long x = getphi(i), y = getphi(n/i);
ans += x * (n/i) + y * i;
if(i * i == n) ans -= x * i;
}
printf("%lld\n",ans);
}
return 0;
}