GCD - Extreme (II) wu-kan

题目链接

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

解法一:蓝书解法,计算贡献

设$f(n)=\sum_{i=1}^{n-1}\gcd(i,n)$,则所求答案为 f(n)的前缀和。 注意到$\gcd(i,n)=k$的值一定是 n 的约数,按照这个约数进行分类。设满足$\gcd(x,n)=k$的约束有 g(n,i)个,则$f(n)=sum\lbrace i\times g(n,i)|d\mod i==0\rbrace $。 依次计算 f(n)需要枚举每个 n 的约数 i,速度较慢,但是对每个 i 枚举它的倍数 ki 来更新 f(ki)的值,此时速度和普通筛法同阶。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct EulerSieve
{
	vector<int> p, m, phi;
	EulerSieve(int N) : m(N, 0), phi(N, 0)
	{
		phi[1] = 1;
		for (long long i = 2, k; i < N; ++i)
		{
			if (!m[i])
				p.push_back(m[i] = i), phi[i] = i - 1;
			for (int j = 0; j < p.size() && (k = i * p[j]) < N; ++j)
			{
				phi[k] = phi[i] * p[j];
				if ((m[k] = p[j]) == m[i])
					break;
				phi[k] -= phi[i];
			}
		}
	}
};
struct GCDExtremeII : EulerSieve
{
	vector<ll> f;
	GCDExtremeII(int N) : EulerSieve(N), f(N)
	{
		for (int i = 1; i < N; ++i)
		{
			for (int k = 2; k * i < N; ++k)
				f[k * i] += i * phi[k];
			f[i] += f[i - 1];
		}
	}
} g(4e6 + 9);
int main()
{
	for (int n; ~scanf("%d", &n) && n;)
		printf("%lld\n", g.f[n]);
}

解法二:分块

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct EulerSieve
{
	vector<int> p, m, phi;
	vector<ll> sum;
	EulerSieve(int N) : m(N, 0), phi(N, 0), sum(N, 0)
	{
		phi[1] = 1;
		for (long long i = 2, k; i < N; ++i)
		{
			if (!m[i])
				p.push_back(m[i] = i), phi[i] = i - 1;
			for (int j = 0; j < p.size() && (k = i * p[j]) < N; ++j)
			{
				phi[k] = phi[i] * p[j];
				if ((m[k] = p[j]) == m[i])
					break;
				phi[k] -= phi[i];
			}
		}
		for (int i = 0; i < N; ++i)
			sum[i] = sum[i - 1] + phi[i];
	}
} e(4e6 + 9);
int main()
{
	for (ll n, sum, tmp; ~scanf("%lld", &n) && n;)
	{
		for (int i = 1, j = sum = 0; i <= n; i = j + 1)
			tmp = n / i, sum += tmp * tmp * (e.sum[j = n / tmp] - e.sum[i - 1]);
		printf("%lld\n", (sum - (n + 1) * n / 2) / 2);
	}
}

解法三:莫比乌斯反演

$ans=\sum_{d=1}^nd\sum_{i=1}^{n/d}\mu(i)[\frac{n}{id}]^2=\sum_{T=1}^n[\frac{n}{T}]^2\sum_{d|T}d\mu(\frac{T}{d})$ 数论分块可$O(\sqrt n)$查询,总时间复杂度$O(n\sqrt n)$。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct EulerSieve
{
	vector<int> p, m, mu, sum; //素数序列,最小素因子,欧拉函数,莫比乌斯函数
	EulerSieve(int N) : m(N, 0), mu(N, 0), sum(N, 0)
	{
		mu[1] = 1;							 //m[1]=0
		for (long long i = 2, k; i < N; ++i) //防i*p[j]爆int
		{
			if (!m[i])
				p.push_back(m[i] = i), mu[i] = -1; //i是素数
			for (int j = 0; j < p.size() && (k = i * p[j]) < N; ++j)
			{
				if ((m[k] = p[j]) == m[i])
				{
					mu[k] = 0;
					break;
				}
				mu[k] = -mu[i];
			}
		}
		for (int i = 1; i < N; ++i)
			sum[i] = sum[i - 1] + mu[i];
	}
} e(4e6 + 9);
int main()
{
	for (ll n, sum, tmp; ~scanf("%lld", &n) && n;)
	{
		sum = 0;
		for (int i = 1; i <= n; ++i)
			for (ll l = 1, r, tmp, m = n / i; l <= m; l = r + 1)
				tmp = m / l, sum += i * tmp * tmp * (e.sum[r = m / tmp] - e.sum[l - 1]);
		printf("%lld\n", (sum - (n + 1) * n / 2) / 2);
	}
}