Luogu2398 GCD SUM

Luogu2398 GCD SUM

求 (displaystylesum_{i=1}^nsum_{j=1}^ngcd(i,j))

(nleq10^5)

数论

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先常规化式子（大雾

[egin{aligned}&displaystylesum_{i=1}^nsum_{j=1}^ngcd(i,j)\&=displaystylesum_{k=1}^nsum_{i=1}^nsum_{j=1}^n{ k imes[gcd(i,j)=k]}\&=displaystylesum_{k=1}^nksum_{i=1}^nsum_{j=1}^n{[gcd(i, j)=k]}\&=displaystylesum_{k=1}^n{ksum_{i=1}^{lfloorfrac{n}{k} floor}}sum_{j=1}^{lfloorfrac{n}{k} floor}{[gcd(i,j)=1]}end{aligned} ]

似乎原问题就转化为了快速求 (displaystyle{sum_{i=1}^{lfloorfrac{n}{k} floor}}sum_{j=1}^{lfloorfrac{n}{k} floor}{[gcd(i,j)=1]})

是不是有点 似曾相识

容易发现上式可以用 (varphi) 替代，即 (2displaystylesum_{i=1}^{lfloorfrac{n}{k} floor}varphi(i)-1)

因此原式即为 $$displaystylesum_{k=1}^{n{k (2displaystylesum_{i=1}}{lfloorfrac{n}{k} floor}varphi(i)-1)}$$

时间复杂度 (O(n)) ，也可以用数论分块优化到 (O(sqrt n))

代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 1e5 + 10;
int n, tot, p[maxn]; ll phi[maxn];

void sieve() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!p[i]) p[++tot] = i, phi[i] = i - 1;
    for (int j = 1; j <= tot && i * p[j] <= n; j++) {
      p[i * p[j]] = 1;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j]; break;
      }
      phi[i * p[j]] = phi[i] * phi[p[j]];
    }
  }
  for (int i = 1; i <= n; i++) {
    phi[i] += phi[i - 1];
  }
}

int main() {
  scanf("%d", &n), sieve(); ll ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += i * (phi[n / i] * 2 - 1);
  }
  printf("%lld", ans);
  return 0;
}