欧拉心算

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n\varphi (gcd(i,j)) \\ =\sum _{d=1}^n\varphi (d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}[gcd(i,j)==1] \\ =\sum _{d=1}^n\varphi (d)\sum _{k=1}^{\frac{n}{d}}\mu (k)(\lfloor \frac{n}{kd}\rfloor {)}^2 \\ 另t=kd \\ =\sum _{t=1}^n(\lfloor \frac{n}{t}\rfloor {)}^2\sum _{d\mid t}\varphi (d)\mu (\frac{t}{d}) \\ 另f(n)=\sum _{d\mid n}\varphi (d)\mu (\frac{n}{d}) \\ 我们考虑如何得到这个函数的前缀和，显然这是一个积性函数有如下性质 \\ f(1)=1 \\ f(p)=\varphi (1)\mu (p)+\varphi (p)\mu (1)=-1+p-1=p-2 \\ 否则我们设n=p^{k}t,p,t互质 \\ f(n)=f(p^k)f(t)=(\sum _{d\mid p^k}\mu (d)\varphi (\frac{p^k}{d}))f(t) \\ =(\sum _{i=0}^k\mu (p^k)\varphi (p^{k-i}))f(t) \\ =(\mu (1)\varphi (p^k)+\mu (p)\varphi (p^{k-1}))f(t) \\ =(p^{k-2}(p^2-2p+1))f(t) \\ 由此我们得到了一个线性筛法,接下来数论分块即可。 \end{gathered}$$

代码

bzoj挂了，所以没地方测试，所以只能在网上找了题解对拍了几组样例。

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;const int N = 1e7 + 10;bool st[N];ll f[N], prime[N], cnt;ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a;a = a * a;n >>= 1;}return ans;
}void init() {st[1] = f[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;f[i] = i - 2;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {int num = 1, temp = i;while(temp % prime[j] == 0) {temp /= prime[j], num++;}f[i * prime[j]] = 1ll * quick_pow(prime[j], num - 2) * (1ll * prime[j] * prime[j] - 2ll * prime[j] + 1) * f[temp];break;}f[i * prime[j]] = f[i] * f[prime[j]];}}for(int i = 1; i < N; i++) {f[i] += f[i - 1];}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while(T--) {ll n, ans = 0;scanf("%lld", &n);for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans += (n / l) * (n / l) * (f[r] - f[l - 1]);}printf("%lld\n", ans);}return 0;
}