[HDU5382]GCD?LCM!

Description

HDU5382

会吗？不会！

设\(F(n)=\sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j)\ge n]\)，求\(S(n)=\sum\limits_{i=1}^{n}F(n)\)

Soluiton

\[F(n) = n^2 - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) < n]\\
F(n) - F(n-1) = n^2 - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) < n] - (n-1)^2 - \sum\limits_{i = 1}^{n-1}\sum\limits_{j=1}^{n-1}[lcm(i,j)+gcd(i,j) < n-1]\\
= 2n - 1 - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n]\\
F(n) = F(n-1) + (2n-1) - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n]
\]

设

\[G(n) = \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n]\\
= \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[\dfrac{k_igcd(i,j)\cdot k_j gcd(i,j)}{gcd(i,j)}+gcd(i,j) = n]\\
= \sum\limits_{d=1}^{n}\sum\limits_{i=1}^{\lfloor \dfrac{n}{d} \rfloor}\sum\limits_{j=1}^{\lfloor \dfrac{n}{d} \rfloor}[ijd + d = n][gcd(i,j) = 1]\\
= \sum\limits_{d|n}\sum\limits_{i=1}^{\dfrac{n}{d}}\sum\limits_{j=1}^{\dfrac{n}{d}}[(ij) = \dfrac{n}{d} - 1][gcd(i,j) = 1]\\
\]

设

\[H(n) = \sum\limits_{i=1}^{n+1}\sum\limits_{j=1}^{n+1}[ij = n][gcd(i,j) = 1]\\
= \sum\limits_{i=1} [gcd(i, \dfrac{n}{i}) = 1]
\]

则

\[G(n) = \sum\limits_{d|n} H(\dfrac{n}{d} - 1)
\]

不难想到，将\(n\)质因数分解后，\(p_x^{a_x}\)要么在\(i\)那一部分，要么在\(\dfrac{n}{i}\)那一部分，所以

\[H(n) = 2^k (k \mbox{ is the number of prime factors of }n)
\]

所以\(H(n)\) 是一个积性函数，可以欧拉筛。然后在计算每个\(H\)对\(G\)的贡献，这样复杂度是\(O(nlogn)\)的，然后就能\(O(n)\)的求出\(F\)和\(S\)。

综上，这个题不涉及NOIp以外的知识，NOIp可以考这么难的

Code

#include <bits/stdc++.h>

typedef long long LL;
const int N = 1e6 + 10;
const LL MOD = 258280327;

LL F[N], G[N], H[N], S[N];
int notp[N], pri[N], cnt;

int get_prime() {
	for (int i = 1; i < N; ++i) H[i] = 1;
	for (int i = 2; i < N; ++i) {
		if (!notp[i]) {
			pri[cnt++] = i;
			H[i] = 2;
		}
		for (int j = 0; j < cnt; ++j) {
			int k = i * pri[j];
			if (k >= N) break;
			notp[k] = 1;
			if (i % pri[j] == 0) {
				(H[k] *= H[i]) %= MOD;
				break;
			}
			else {
				(H[k] *= 2 * H[i] % MOD) %= MOD;
			}
		}
	}
	for (int i = 1; i < N; ++i) {
		for (int j = i; j < N; j += i) {
			G[j] = (G[j] + H[j/i - 1]) % MOD;
		}
	}
	F[1] = 1;
	for (int i = 2; i < N; ++i) {
		F[i] = ((LL)F[i-1] + i + i - 1LL - G[i-1]) % MOD;
	}
	for (int i = 1; i < N; ++i) {
		S[i] = (S[i-1] + F[i]) % MOD;
	}
}

int main() {
	get_prime();
	int t;
	scanf("%d", &t);
	while (t--) {
	    int n;
	    scanf("%d", &n);
	    printf("%d\n", S[n]);
	}
	return 0;
}