Luogu2398 GCD SUM

Luogu2398 GCD SUM

求 \(\displaystyle\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j)\)

\(n\leq10^5\)

数论

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先常规化式子（大雾

\[\begin{aligned}&\displaystyle\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j)\\&=\displaystyle\sum_{k=1}^n\sum_{i=1}^n\sum_{j=1}^n{\ k\times[\gcd(i,j)=k]}\\&=\displaystyle\sum_{k=1}^nk\sum_{i=1}^n\sum_{j=1}^n{[\gcd(i, j)=k]}\\&=\displaystyle\sum_{k=1}^n{k\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}}\sum_{j=1}^{\lfloor\frac{n}{k}\rfloor}{[\gcd(i,j)=1]}\end{aligned} \]

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

是不是有点 似曾相识

容易发现上式可以用 \(\varphi\) 替代，即 \(2\displaystyle\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\varphi(i)-1\)

因此原式即为 $$\displaystyle\sum_{k=1}^{n{k (2\displaystyle\sum_{i=1}}\rfloor}\varphi(i)-1)}$$

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;
}