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#include <bits/stdc++.h>
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using namespace std;
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#define int long long
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const int N = 1e6 + 10;
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// 筛法求欧拉函数
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// 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
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int primes[N], phi[N], st[N];
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int cnt, res;
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void get_eulers(int n) {
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phi[1] = 1; // 1的欧拉函数值是1,这个是递推的起点
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for (int i = 2; i <= n; i++) {
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if (!st[i]) {
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primes[cnt++] = i;
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phi[i] = i - 1; // ①质数,则phi(i)=i-1
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}
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for (int j = 0; primes[j] <= n / i; j++) {
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int t = primes[j] * i;
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st[t] = 1;
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if (i % primes[j] == 0) {
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phi[t] = phi[i] * primes[j]; // ② i%pj==0
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break;
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} else
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phi[t] = phi[i] * (primes[j] - 1); // ③i%pj>0
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}
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}
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}
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signed main() {
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int n;
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cin >> n;
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// 筛法求欧拉函数
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get_eulers(n);
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// 累加欧拉函数值
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for (int i = 1; i <= n; i++) res += phi[i];
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printf("%lld\n", res);
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}
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