[HDU5382]GCD?LCM!
阿新 • • 發佈：2018-09-20
.cn otp nlog mbo n) problem splay break rac
設
\[ 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]\\]
設
\[ H(n) = \sum\limits_{i=1}^{n+1}\sum\limits_{j=1}^{n+1}[ij = n][gcd(i,j) = 1]\= \sum\limits_{i=1} [gcd(i, \dfrac{n}{i}) = 1] \]
則
\[ G(n) = \sum\limits_{d|n} H(\dfrac{n}{d} - 1) \]
不難想到，將\(n\)質因數分解後，\(p_x^{a_x}\)要麽在\(i\)那一部分，要麽在\(\dfrac{n}{i}\)那一部分，所以
\[ H(n) = 2^k (k \mbox{ is the number of prime factors of }n) \]
所以\(H(n)\) 是一個積性函數，可以歐拉篩。然後在計算每個\(H\)對\(G\)的貢獻，這樣復雜度是\(O(nlogn)\)的，然後就能\(O(n)\)的求出\(F\)和\(S\)。
Description
HDU5382
會嗎？不會！
設\(F(n)=\sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j)\ge n]\)，求\(S(n)=\sum\limits_{i=1}^{n}F(n)\)
Soluiton
\[ F(n) = n^2 - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) < n]\F(n) - F(n-1) = n^2 - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) < n] - (n-1)^2 - \sum\limits_{i = 1}^{n-1}\sum\limits_{j=1}^{n-1}[lcm(i,j)+gcd(i,j) < n-1]\= 2n - 1 - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n]\F(n) = F(n-1) + (2n-1) - \sum\limits_{i = 1}^{n}\sum\limits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n] \]
設
\[ 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]\\]
設
\[ H(n) = \sum\limits_{i=1}^{n+1}\sum\limits_{j=1}^{n+1}[ij = n][gcd(i,j) = 1]\= \sum\limits_{i=1} [gcd(i, \dfrac{n}{i}) = 1] \]
則
\[ G(n) = \sum\limits_{d|n} H(\dfrac{n}{d} - 1) \]
不難想到，將\(n\)質因數分解後，\(p_x^{a_x}\)要麽在\(i\)那一部分，要麽在\(\dfrac{n}{i}\)那一部分，所以
\[ H(n) = 2^k (k \mbox{ is the number of prime factors of }n) \]
所以\(H(n)\) 是一個積性函數，可以歐拉篩。然後在計算每個\(H\)對\(G\)的貢獻，這樣復雜度是\(O(nlogn)\)的，然後就能\(O(n)\)的求出\(F\)和\(S\)。
綜上，這個題不涉及NOIp以外的知識，NOIp可以考這麽難的
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;
}[HDU5382]GCD?LCM!