积性函数板刷1

A-线性筛求积性函数

求正整数 $n$ 的所有正因数的个数，$q$ 次询问。

第一行一个正整数 $q\ (1\leq q\leq 10^5)$ 。
第二行到第 $q+1$ 行，每行一个正整数 $n\ (1\leq n\leq 10^7)$。

code

// NowCoder 模板题【线性筛求积性函数】
// https://ac.nowcoder.com/acm/contest/22769/A 2024-03-28 21:02:44

#include <bits/stdc++.h>
using namespace std;
#define all(a) begin(a), end(a)
#define int long long
int n, m;
constexpr int N = 1E7 + 10;
int primes[N], tot;
int st[N], f[N];
void init() {
  f[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!st[i]) {
      primes[++tot] = i;
      f[i] = 2;
    }
    for (int j = 1, cnt = 0; j <= tot and i  primes[j] < N; j++) {
      st[i  primes[j]] = true;
      if (i % primes[j] == 0) {
        int w = i;
        while (w % primes[j] == 0) cnt++, w /= primes[j];
        f[i  primes[j]] = f[i] / (cnt + 1)  (cnt + 2);
        break;
      } else
        f[i  primes[j]] = f[i]  2;
    }
  }
}
void solve() {
  cin >> n;
  cout << f[n] << "\n";
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  init();
  int __;
  cin >> __;
  while (__--) solve();
  return 0;
}

C-序列

THINK

F(n)为gcd(x,y)为n的倍数的有序对的对数

f(n)为gcd(x,y)为n的有序对的对数
$$ \begin{align} Ans&=\sum \limits _{1≤x≤n,1≤y≤n}[gcd(x,y)=1]⋅[a_{b_x}=b_{a_y}]\\ F(n)&=\sum\limits _{n|d}f(d)\\ f(n)&=\sum\limits _{n|d}\mu(\frac d n)\times F(d)\\ \end{align} $$
最后求出f(1)

code

// NowCoder 序列
// https://ac.nowcoder.com/acm/contest/22769/C 2024-03-28 21:41:19

#include <bits/stdc++.h>
using namespace std;
#define all(a) begin(a), end(a)
#define int long long
constexpr int N = 1E5 + 10;
int primes[N], tot;
int st[N], mu[N], F[N];
void init(int N) {
  mu[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!st[i]) {
      primes[++tot] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= tot and i * primes[j] <= N; j++) {
      st[i * primes[j]] = true;
      if (i % primes[j] == 0) {
        mu[i * primes[j]] = 0;
        break;
      } else
        mu[i * primes[j]] = -mu[i];
    }
  }
}
int n, m;
void solve() {
  cin >> n;
  init(n);
  vector<int> a(n + 1);
  vector<int> b(n + 1);
  vector<int> cnt(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j += i) {
      cnt[a[b[j]]]++;
    }
    for (int j = i; j <= n; j += i) {
      F[i] += cnt[b[a[j]]];
    }
    for (int j = i; j <= n; j += i) {
      cnt[a[b[j]]]--;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += mu[i] * F[i];
  }
  cout << ans << "\n";
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);

  solve();
  return 0;
}