BZOJ 2226 LCMSum

Description

Given \(n\), calculate the sum \(LCM(1,n) + LCM(2,n) + \cdots + LCM(n,n)\), where \(LCM(i,n)\) denotes the Least Common Multiple of the integers \(i\) and \(n\).

Input

The first line contains \(T\) the number of test cases. Each of the next \(T\) lines contain an integer \(n\).

Output

Output \(T\) lines, one for each test case, containing the required sum.

Sample Input

3

1

2

5

Sample Output

1

4

55

HINT

\(1 \le T \le 300000\)

\(1 \le n \le 1000000\)

题目求$$\sum_{i=1}^{n}LCM(i,n)$$

根据\(LCM\)的公式，即$$\sum_{i=1}^{n}\frac{i \times n}{GCD(i,n)}$$

我们枚举\(GCD\)——\(g\)，即$$\sum_{g=1}^{n}[g \mid n]n \sum_{i=1}^{n}i[GCD(i,n)=g]$$

化简一下，转而求$$\sum_{g=1}^{n}[g \mid n]n \sum_{i=1}^{n}i[GCD(\frac{i}{g},\frac{n}{g})=1]$$

变化一下\(i\)的范围：$$\sum_{g=1}^{n}[g \mid n]n \sum_{i=1}^{\frac{n}{g}}i[GCD(i,\frac{n}{g})=1]$$

\(\sum_{i=1}^{\frac{n}{g}}i[GCD(i,\frac{n}{g})=1]\)即\(\frac{n}{g}\)内与之互质的数的和，这个有个公式：$$\sum_{i=1}^{n}i[GCD(n,i)=1]= \frac{\phi(n) \times n}{2}$$

如何证明，假设某个数\(a\)与\(n\)互质，那么\(n-a\)一定也与\(n\)互质，这样的数一共有\(\phi(n)\)个，于是得证，但在\(n=1\)是要特判，于是这个式子就出来了。$$\sum_{i=1}^{{n}LCM(i,n)=\sum_{g=1}}{n}[g \mid n]n \frac{\phi(\frac{n}{g}) \times \frac{n}{g} }{2}$$

#include<cstdio>
#include<cstdlib>
using namespace std;

typedef long long ll;
#define maxn (1000010)
bool exist[maxn]; int n,phi[maxn],prime[maxn],tot;

inline void ready()
{
	phi[1] = 1;
	for (int i = 2;i < maxn;++i)
	{
		if (!exist[i]) phi[i] = i-1,prime[++tot] = i;
		for (int j = 1;j <= tot;++j)
		{
			if (i*prime[j] >= maxn) break;
			exist[i*prime[j]] = true;
			if (i % prime[j] == 0) { phi[i*prime[j]] = phi[i]*prime[j]; break; }
			else phi[i*prime[j]] = phi[i]*phi[prime[j]];
		}
	}
}
inline ll calc(int g)
{
	if (g == 1) return 1;
	return ((ll)phi[g]*(ll)g>>1);
}

int main()
{
	freopen("2226.in","r",stdin);
	freopen("2226.out","w",stdout);
	ready();
	int T; scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n);
		ll ans = 0;
		for (int g = 1;g * g <= n;++g)
			if (n % g == 0)
			{
				ans += (ll)n*calc(n / g);
				if (g * g != n) ans += (ll)n*calc(g);
			}
		printf("%lld\n",ans);
	}
	fclose(stdin); fclose(stdout);
	return 0;
}