题目传送门

题意：

给你一个 n ( 1 ≤ n ≤ 2 ∗ 1 0 6 ) n(1\le n\le 2*10^6) n(1≤n≤2∗106)，求 ∑ i = 1 n ∑ j = i + 1 n g c d ( i , j ) \sum\limits_{i=1}^{n}\sum\limits_{j=i+1}^{n}gcd(i,j) i=1∑n​j=i+1∑n​gcd(i,j)。

思路：

稍微对这个式子改写一下，变成了 ∑ i = 1 n ∑ j = 1 n g c d ( i , j ) − ( 1 + n ) ∗ n 2 2 \cfrac{\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}gcd(i,j)-\cfrac{(1+n)*n}{2}}{2} 2i=1∑n​j=1∑n​gcd(i,j)−2(1+n)∗n​​。

我们只研究式子 ∑ i = 1 n ∑ j = 1 n g c d ( i , j ) \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}gcd(i,j) i=1∑n​j=1∑n​gcd(i,j)就行了。

∑ i = 1 n ∑ j = 1 n g c d ( i , j ) = ∑ g = 1 n ∑ i = 1 n ∑ j = 1 n [ g c d ( i , j ) = g ] ∗ g = ∑ g = 1 n ∑ i = 1 ⌊ n g ⌋ ∑ j = 1 ⌊ n g ⌋ [ g c d ( i , j ) = 1 ] ∗ g = ∑ g = 1 n ∑ i = 1 ⌊ n g ⌋ ∑ j = 1 ⌊ n g ⌋ ∑ d ∣ ( i , j ) μ ( d ) ∗ g = ∑ g = 1 n ∑ d ∣ ( i , j ) μ ( d ) ⌊ n g d ⌋ 2 ∗ g \begin{aligned} \\ &\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}gcd(i,j)\\ =&\sum\limits_{g=1}^{n}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}[gcd(i,j)=g]*g\\ =&\sum\limits_{g=1}^{n}\sum\limits_{i=1}^{\lfloor \frac{n}{g}\rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{g}\rfloor}[gcd(i,j)=1]*g\\ =&\sum\limits_{g=1}^{n}\sum\limits_{i=1}^{\lfloor \frac{n}{g}\rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{g}\rfloor}\sum\limits_{d|(i,j)}\mu(d)*g\\ =&\sum\limits_{g=1}^{n}\sum\limits_{d|(i,j)}\mu(d)\lfloor\cfrac{n}{gd}\rfloor^2*g \end{aligned} ====​i=1∑n​j=1∑n​gcd(i,j)g=1∑n​i=1∑n​j=1∑n​[gcd(i,j)=g]∗gg=1∑n​i=1∑⌊gn​⌋​j=1∑⌊gn​⌋​[gcd(i,j)=1]∗gg=1∑n​i=1∑⌊gn​⌋​j=1∑⌊gn​⌋​d∣(i,j)∑​μ(d)∗gg=1∑n​d∣(i,j)∑​μ(d)⌊gdn​⌋2∗g​

然后就可以整除分块求答案了。

C o d e Code Code

// Author : ACfunhsl
// Time : 2021/5/17 14:13:11
#define int long long
const int N = 2e6+50;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
bool ok[N];
int p[N],cnt=0,mu[N];
void euler()
{
	mu[1] = 1;
	for(int i=2; i<N; i++)
	{
		if(!ok[i])
		{
			p[++cnt] = i;
			mu[i] = -1;
		}
		for(int j=1; j<=cnt&&i*p[j]<N; j++)
		{
			ok[i*p[j]] = 1;
			if(i%p[j]==0) break;
			mu[i*p[j]] = -mu[i];
		}
	}
	for(int i=1; i<N; i++)
		mu[i] += mu[i-1];
}
int cal(int n,int k)
{
	int res = 0;
	n/=k;
	for(int l=1,r; l<=n; l=r+1)
	{
		r = n/(n/l);
		res += (n/l)*(n/l)*(mu[r] - mu[l-1]);
	}
	return res;
}
signed main()
{
	euler();
	int n;
	cin>>n;
	int res = 0;
	for(int i=1;i<=n;i++)
		res += cal(n,i)*i;
	res -= (1+n)*n/2;
	res /= 2;
	cout<<res<<endl;
	

	return 0;
}