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牛客 contest 43058E 炫酷反演魔术

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题目描述

https://ac.nowcoder.com/acm/contest/43058/E

简要题意：给定一个长度为 $n$ 的序列 $a_i$，求 $\sum_{i=1}^n\sum_{j=1}^n\varphi(gcd(a_i,a_j^3))$

$n\le 3\times 10^5,a_i\in[1,n]$

Solution

我们直接化简

注意到我们只需要预处理 $\sum_{d|a_i}$ 和 $\sum_{d|a_i^3}$ 即可，后面那个东西本质和前面的一样，时间复杂度 $O(n\log n)$

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#include <iostream>
#include <algorithm>
#include <queue>
#define maxn 300010
#define ll long long
#define INF 1000000000
using namespace std;

bool isp[maxn]; int mu[maxn], pri[maxn], phi[maxn], a[maxn];
void init_isp(int n) {
    phi[1] = mu[1] = 1; int cnt = 0; 
    for (int i = 2; i <= n; ++i) {
        if (!isp[i]) pri[++cnt] = i, phi[i] = i - 1, a[i] = i, mu[i] = -1;
        for (int j = 1; j <= cnt && i * pri[j] <= n; ++j) {
            isp[i * pri[j]] = 1;
            a[i * pri[j]] = pri[j];
            if (i % pri[j] == 0) {
                phi[i * pri[j]] = phi[i] * pri[j];
                break;
            }
            phi[i * pri[j]] = phi[i] * (pri[j] - 1);
            mu[i * pri[j]] = -mu[i];
        }
    } 
}

int n, cnt[maxn], c1[maxn], c2[maxn], t[maxn];

vector<int> A[maxn], B[maxn];

int main() { 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n; init_isp(n);
    for (int i = 1, x; i <= n; ++i) cin >> x, ++cnt[x];
    for (int i = 1; i <= n; ++i) {
        int x = i, y = 1; 
        while (x != 1) {
            int p = a[x], t = 0;
            while (x % p == 0) ++t, x /= p;
            for (int i = 1; i <= (t + 2) / 3; ++i) y *= p;
        } t[i] = y; 
    }
    for (int i = 1; i <= n; ++i)
        for (int j = i; j <= n; j += i) c1[i] += cnt[j];
    for (int i = 1; i <= n; ++i) c2[i] = c1[t[i]];
    ll ans = 0; 
    for (int i = 1; i <= n; ++i)
        for (int j = i; j <= n; j += i) ans += 1ll * phi[i] * mu[j / i] * c1[j] * c2[j];
    cout << ans << "\n";
    return 0;
}