代码

#include <iostream>

using namespace std;

const int N = 1e6 + 10;

typedef long long LL;

int n;
int primes[N], cnt;
int phi[N];
bool st[N];

LL get_eulers(int n)
{
    phi[1] = 1;  // 从 1 ~ 1， 与 1 互质的数的个数为 1
    for (int i = 2; i <= n; i++) {
        if (!st[i]) {
            primes[cnt++] = i;
            phi[i] = i - 1;
        }
        for (int j = 0; primes[j] <= n / i; j++) {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) {
                phi[i * primes[j]] = phi[i] * primes[j];
                break;
            }
            //如果 i%primes[j]!=0
            phi[i * primes[j]] = phi[i] * (primes[j] - 1);
        }
    }
    LL res = 0;
    for (int i = 1; i <= n; i++) {
        res+=phi[i];
    }
    return res;
}

int main()
{
    cin >> n;
    cout << get_eulers(n) << endl;
    return 0;
}