連接
推一下就是\(\sum_{k=1}^{n}\lfloor\frac{n}{k}\rfloor^2\sum_{d|k}\phi(d)\mu(\frac{k}{d})\)
\(\sum_{d|k}\phi(d)\mu(\frac{k}{d})\)線性篩一下就好php
# include <bits/stdc++.h> # define RG register # define IL inline # define Fill(a, b) memset(a, b, sizeof(a)) using namespace std; typedef long long ll; const int _(1e7 + 1); IL ll Read(){ RG ll x = 0, z = 1; RG char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); return x * z; } int prime[_], num; ll f[_]; bool isprime[_]; IL void Prepare(){ isprime[1] = 1; f[1] = 1; for(RG int i = 2; i < _; ++i){ if(!isprime[i]){ prime[++num] = i; f[i] = i - 2; } for(RG int j = 1; j <= num && i * prime[j] < _; ++j){ isprime[i * prime[j]] = 1; if(i % prime[j]) f[i * prime[j]] = f[i] * f[prime[j]]; else{ if((i / prime[j]) % prime[j]) f[i * prime[j]] = f[i / prime[j]] * (prime[j] - 1) * (prime[j] - 1); else f[i * prime[j]] = f[i] * prime[j]; break; } } } for(RG int i = 2; i < _; ++i) f[i] += f[i - 1]; } int main(RG int argc, RG char* argv[]){ Prepare(); for(RG ll T = Read(), n, ans; T; --T){ n = Read(); ans = 0; for(RG ll i = 1, j; i <= n; i = j + 1){ j = n / (n / i); ans += (n / i) * (n / i) * (f[j] - f[i - 1]); } printf("%lld\n", ans); } return 0; }