题意
Sol
开始用反演推发现不会求\(\mu(k)\)慌的一批
退了两步发现只要求个欧拉函数就行了
\(ans = \sum_{d | n} d \phi(\frac{n}{d})\)
理论上来说复杂度是\(O(n)\)的，但是\(d\)的值十分有限。在\(2^{32}\)内最多的约数也只有1920个。
/*
*/
#include<bits/stdc++.h>
#define LL long long
#define int long long
const int MAXN = 1e5 + 10, INF = 1e9 + 7;
using namespace std;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N;
int calc(int N) {
int res = 1;
for(int i = 2; i * i <= N; i++) {
if(N % i == 0) {
int now = (i - 1); N /= i;
while(N % i == 0) now *= i, N /= i;
res *= now;
}
}
if(N != 1) res *= (N - 1);
return res;
}
signed main() {
N = read();
int ans = 0;
for(int i = 1; i * i <= N; i++) {
if(N % i == 0) {
ans += i * calc(N / i);
if(i != N / i) ans += (N / i) * calc(i);
}
}
cout << ans;
return 0;
}
/*
3 7
a*b
aebr*ob
*/