题目：求 $\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j)$ 的值。

推式子

$$\sum _{i=1}^n\sum _{j=1}^n\gcd {\left(i,j\right)}^2$$
$$=\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n{k}^2\times \left[\gcd (i,j)==k\right]$$
$$=\sum _{k=1}^n{k}^2\sum _{i=1}^n\sum _{j=1}^n\gcd (i,j)$$
$$=\sum _{k=1}^n{k}^2\left(2\sum _{i=1}^{\lfloor \frac{n}{k}\rfloor }\phi (i)-1\right)$$

代入杜教筛公式。
$$=\sum _{k=1}^n{k}^2\left(n^2+n-\sum _{i=2}^{n}S\left(\lfloor \frac{n}{i}\rfloor \right)\right)$$

实现

先用欧拉筛筛出前 $K$ 个，再求。
复杂度 $\Theta \left(K+\frac{n}{\sqrt{K}}\right)=\Theta (n^{\frac{2}{3}})$.

代码:

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1000000007;
unordered_map<int, int>mp;
int K = 10000;
int s[500010];
inline int S(int x) {if (x <= K)return s[x];//预处理if (mp.find(x) != mp.end())return mp[x];//记忆化int j = 0;int ans = 1ll * (x) * (x + 1) / 2;for (int i = 2; i <= x; i = j + 1) {j = x / (x / i);//数论分块ans -= (j - i + 1) * S(x / i);//个数*每个的值}return mp[x] = (ans % mod);
}
int primes[500010];
int cnt;
bool st[500010];
inline void init() {//欧拉筛求欧拉函数s[1] = 1;for (int i = 2; i <= K; i++) {if (!st[i]) {primes[cnt++] = i;s[i] = i - 1; //质数前面所有数都互质}for (int j = 0; primes[j] <= K / i; j++) {int t = primes[j] * i;st[t] = true;if (i % primes[j] == 0) {s[t] = primes[j] * s[i];break;}s[t] = s[i] * (primes[j] - 1);}}for (int i = 1; i <= K; i++)s[i] += s[i - 1], s[i] %= mod;
}
signed main() {int n;cin >> n;init();int ans = 0;for (int i = 1; i <= n; ++i)ans = (ans + ((i * i) % mod * ( (S(n / i) << 1) % mod - 1 )) % mod) % mod;//公式cout << ans;return 0;
}