Luogu P1390 公约数的和

gate
求：

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^m gcd(i,j)$$

设$gcd(i,j) = k$，枚举$k$

$$\sum\limits_{k=1}^n k \sum\limits_{i=1}^n\sum\limits_{j=1}^m [gcd(i,j)=k]$$

同时除以$k$

$$\sum\limits_{k=1}^n k \sum\limits_{i=1}^{\lfloor \frac{n}{k}\rfloor}\sum\limits_{j=1}^{\lfloor \frac{m}{k}\rfloor} [gcd(i,j)=1]$$

根据
$\begin{array} \ \because \sum\limits_{d∣n}\mu(d)=[n=1] \\ \therefore \sum\limits_{d∣gcd(i,j)}​μ(d) = [gcd(i,j)=1] \end{array}$
把$[gcd(i,j)=1]$替换掉

$$\sum\limits_{k=1}^n k \sum\limits_{i=1}^{\lfloor \frac{n}{k}\rfloor}\sum\limits_{j=1}^{\lfloor \frac{m}{k}\rfloor} \sum\limits_{d∣n}\mu(d)$$

将$d$提前，同时把$i,j$除以$d$

$$\sum\limits_{k=1}^n k\sum\limits_{d=1}^{\lfloor \frac{n}{k}\rfloor}\sum\limits_{i=1}^{\lfloor \frac{n}{kd}\rfloor}\sum\limits_{j=1}^{\lfloor \frac{m}{kd}\rfloor}\mu(d)$$

整除分块

$$\sum\limits_{k=1}^n k\sum\limits_{d=1}^{\lfloor \frac{n}{k}\rfloor}\mu(d)\lfloor \frac{n}{kd}\rfloor\lfloor \frac{m}{kd}\rfloor$$

设$T=kd$

$$\sum\limits_{k=1}^n k\sum\limits_{d=1}^{\lfloor \frac{n}{k}\rfloor}\mu(d)\lfloor \frac{n}{T}\rfloor\lfloor \frac{m}{T}\rfloor$$

把$d$换成$T$

$$\sum\limits_{T=1}^n\sum\limits_{k|T} k\mu(\frac{T}{k})\lfloor \frac{n}{T}\rfloor\lfloor \frac{m}{T}\rfloor$$

$k$也就是$id(k)$，可以写成

$$\sum\limits_{T=1}^n\sum\limits_{k|T} id(k)\mu(\frac{T}{k})\lfloor \frac{n}{T}\rfloor\lfloor \frac{m}{T}\rfloor$$

因为$k*\frac{T}{k} = T$，所以在枚举$T$时，$k$和$\frac{T}{k}$可以看成相同的元素，只是顺序相反
$\begin{array} \ \because id*\mu=\varphi \\ \therefore \sum\limits_{k|T}id(k)\mu(\frac{T}{k}) = \sum\limits_{k|T}\varphi(k) \end{array}$
$\because \varphi$是积性函数
$\therefore \sum\limits_{k|T}\varphi(k) = \varphi(T)$
即原式可以转化为

$$\sum\limits_{T=1}^n\varphi(T)\lfloor \frac{n}{T}\rfloor\lfloor \frac{m}{T}\rfloor$$

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莫比乌斯反演的部分到此结束了，但注意，题目要求的是$[1,n]$中每任意两个不同的数的$gcd$之和，
而上述做法多加了$gcd(i,i)$，且重复计算了$gcd(i,j)$和$gcd(j,i)$
所以我们要把多余的部分减去，最终答案即 $(Ans-\sum\limits_{i=1}^ni)/2$
也就是等差数列，可以写成$(Ans-(1+n)*n/2)/2$

代码如下

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define MogeKo qwq
using namespace std;

const int maxn = 2e6+5;
long long n,p[maxn];

void get_phi(long long n) {
	for(long long i = 1; i <= n; i++)
		p[i] = i;
	for(long long i = 2; i <= n; i++)
		if(p[i] == i)
			for(long long j = i; j <= n; j += i)
				p[j] = p[j]/i*(i-1);
	for(long long i = 1; i <= n; i++)
		p[i] += p[i-1];
}

long long calc(long long n,long long m) {
	long long ans = 0;
	long long r = 0;
	for(long long i = 1; i <= n; i = r+1) {
		r = min(n/(n/i),m/(m/i));
		ans += (p[r]-p[i-1]) * (n/i) * (m/i);
	}
	return ans;
}

int main() {
	scanf("%lld",&n);
	get_phi(n);
	printf("%lld",(calc(n,n)-(1+n)*n/2)/2);
	return 0;
}