\[\texttt{Description} \]
给出 \(n\)，求 \(\sum\limits_{i = 1}^n\sum\limits_{j=i+1}^n \gcd(i,j)\)。
\[\texttt{Solution} \]
注意到：
\[\sum\limits_{i = 1}^n\sum\limits_{j = i + 1}^n \gcd(i, j) = \frac{\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n \gcd(i, j) - \sum\limits_{i = 1}^n \gcd(i,i)}{2} = \frac{\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n \gcd(i, j) - \frac{n \cdot (n + 1)}{2}}{2} \]
于是问题转化为如何求出 \(\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n \gcd(i, j)\)。
考虑枚举 \(d = \gcd(i, j)\)，则式子化为：
\[\sum\limits_{d = 1}^n d \cdot \sum\limits_{i = 1}^n\sum\limits_{j = 1}^n [\gcd(i,j) = d] \]
\[\sum\limits_{d = 1}^n d \cdot \sum\limits_{i = 1}^{\frac{n}{d}}\sum\limits_{j = 1}^{\frac{n}{d}} [\gcd(i,j) = 1] \]
设 \(g(n) = \sum\limits_{i = 1}^n\sum\limits_{j = 1}^n [\gcd(i,j) = 1]\)，则式子化为：
\[\sum\limits_{d = 1} ^ n d \cdot g(\frac{n}{d}) \]
考虑化简 \(g(n)\)，注意到：
\[\varphi(n) = \sum\limits_{i = 1}^n [\gcd(n, i) = 1] \]
则有：
\[g(n) = 2 \cdot \sum\limits_{i = 1}^n \varphi(i) - 1 \]
线性筛 \(\mathcal{O(n)}\) 预处理出欧拉函数，再求一遍欧拉函数前缀和，即可 \(\mathcal{O(1)}\) 计算 \(g(n)\)。
至此 \(\sum\limits_{d = 1} ^ n d \cdot g(\frac{n}{d})\) 已经可以 \(\mathcal{O(n)}\) 计算，可以进一步用 " 数论分块 " 优化到 \(O(\sqrt n)\)。
\[\texttt{Code} \]
#include
using namespace std;
const int N = 2000100;
int n;
int m, prime[N], fac[N];
int phi[N];
long long S[N];
void GetPrimes(int N) {
phi[1] = 1;
for (int i = 2; i <= N; i ++) {
if (!fac[i]) fac[i] = i, prime[++ m] = i, phi[i] = i - 1;
for (int j = 1; j <= m; j ++) {
if (prime[j] > fac[i] || prime[j] > N / i) break;
fac[i * prime[j]] = prime[j];
phi[i * prime[j]] = phi[i] * (i % prime[j] ? prime[j] - 1 : prime[j]);
}
}
for (int i = 1; i <= N; i ++)
S[i] = S[i - 1] + phi[i];
}
long long g(int n) {
return 2 * S[n] - 1;
}
int main() {
GetPrimes(2000000);
scanf("%d", &n);
long long ans = 0;
for (int x = 1, Nx; x <= n; x = Nx + 1) {
Nx = n / (n / x);
ans += 1ll * (x + Nx) * (Nx - x + 1) / 2 * g(n / x);
}
ans -= 1ll * n * (n + 1) / 2;
ans /= 2;
printf("%lld\n", ans);
return 0;
}
\[\texttt{Thanks} \ \texttt{for} \ \texttt{reading} \]