A Simple Math Problem
/* Author : lifehappy */ #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e5 + 10; ll prime[N], phi[N], mu[N], f[N], g[N], cnt, n; bool st[N]; void init() { phi[1] = mu[1] = f[1] = 1; for(int i = 2; i <= n; i++) { f[i] = f[i / 10] + i % 10; if(!st[i]) { prime[++cnt] = i; phi[i] = i - 1; mu[i] = -1; } for(int j = 1; j <= cnt && 1ll * i * prime[j] <= n; j++) { st[i * prime[j]] = 1; if(i % prime[j] == 0) { phi[i * prime[j]] = phi[i] * prime[j]; break; } phi[i * prime[j]] = phi[i] * (prime[j] - 1); mu[i * prime[j]] = -mu[i]; } } for(int i = 1; i <= n; i++) { for(int j = i; j <= n; j += i) { g[i] += f[j]; } } } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); scanf("%lld", &n); init(); ll ans1 = 0, ans2 = 0; for(int i = 1; i <= n; i++) { ans1 += mu[i] * (n / i) * g[i]; ans2 += f[i] * phi[i]; } printf("%lld\n", ans1 - ans2 + 1); return 0; }