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2 years ago
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
// 筛法求欧拉函数
// 这个东西就在欧兰筛的基础上,加上题解中证明的公式,只能是背下来,没有别的办法
int primes[N], phi[N], st[N];
int cnt, res;
void get_eulers(int n) {
phi[1] = 1; // 1的欧拉函数值是1这个是递推的起点
for (int i = 2; i <= n; i++) {
if (!st[i]) {
primes[cnt++] = i;
phi[i] = i - 1; // ①质数,则phi(i)=i-1
}
for (int j = 0; primes[j] <= n / i; j++) {
int t = primes[j] * i;
st[t] = 1;
if (i % primes[j] == 0) {
phi[t] = phi[i] * primes[j]; // ② i%pj==0
break;
} else
phi[t] = phi[i] * (primes[j] - 1); // ③i%pj>0
}
}
}
signed main() {
int n;
cin >> n;
// 筛法求欧拉函数
get_eulers(n);
// 累加欧拉函数值
for (int i = 1; i <= n; i++) res += phi[i];
printf("%lld\n", res);
}