[HDU5382]GCD?LCM!

Posted 2021-01-07 wyxwyx

Description

HDU5382
会吗？不会！
设(F(n)=sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[lcm(i,j)+gcd(i,j)ge n])，求(S(n)=sumlimits_{i=1}^{n}F(n))

Soluiton

[ F(n) = n^2 - sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[lcm(i,j)+gcd(i,j) < n]F(n) - F(n-1) = n^2 - sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[lcm(i,j)+gcd(i,j) < n] - (n-1)^2 - sumlimits_{i = 1}^{n-1}sumlimits_{j=1}^{n-1}[lcm(i,j)+gcd(i,j) < n-1]= 2n - 1 - sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n]F(n) = F(n-1) + (2n-1) - sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n] ]
设
[ G(n) = sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[lcm(i,j)+gcd(i,j) = n]= sumlimits_{i = 1}^{n}sumlimits_{j=1}^{n}[dfrac{k_igcd(i,j)cdot k_j gcd(i,j)}{gcd(i,j)}+gcd(i,j) = n]= sumlimits_{d=1}^{n}sumlimits_{i=1}^{lfloor dfrac{n}{d} floor}sumlimits_{j=1}^{lfloor dfrac{n}{d} floor}[ijd + d = n][gcd(i,j) = 1]= sumlimits_{d|n}sumlimits_{i=1}^{dfrac{n}{d}}sumlimits_{j=1}^{dfrac{n}{d}}[(ij) = dfrac{n}{d} - 1][gcd(i,j) = 1]\]
设
[ H(n) = sumlimits_{i=1}^{n+1}sumlimits_{j=1}^{n+1}[ij = n][gcd(i,j) = 1]= sumlimits_{i=1} [gcd(i, dfrac{n}{i}) = 1] ]
则
[ G(n) = sumlimits_{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
", S[n]);
    }
    return 0;
}