Luo Gu 2257: YY the GCD

Luo Gu 2257: YY the GCD

Description of the meaning of problems

• Q \ (\ sum_. 1} = {I ^ N \ sum_. 1} = {J ^ M [GCD (I, J) == P] \) , where \ (P \) are all prime numbers.
• Data range \ (N, M \. 7 Leq 10 ^ \) , there are \ (10000 \) set of test data.

Ideas:

• Mobius inversion
• Set \ (f (d) \) of \ (gcd (i, j) == d \) number: \ (F (D) = \ sum_. 1} ^ {n-I = \ sum_ = {J}. 1 m ^ [GCD (I, J) == D] \) .
• \ (F (n) \) of \ (gcd (i, j) == kn \) number, where \ (K = 1,2, ... \) .
• 有\(F(n)=\sum_{n|d}f(d)=\frac{N}{n}\frac{M}{n}\)。

• \(f(n)=\sum_{n|d}\mu(\frac{d}{n})F(d)\)。

• \(ans=\sum_{p\in primes}\sum_{i=1}^N\sum_{j=1}^M[gcd(i,j)==p]\).

• \(=\sum_{p\in primes}f(p)\)。

• \(=\sum_{p\in primes}\sum_{p|d}\mu(\frac{d}{p})F(d)\).

• Enumeration \ (\ FRAC {P} = {D} T \) , there

• \(=\sum_{p\in primes}\sum_{t=1}^{min\{\frac{N}{p},\frac{M}{p}\}}\mu(t)\frac{N}{tp}\frac{M}{tp}\)。

• Set \ (T = TP \) , are:

• \(\sum_{p\in primes}\sum_{t=1}^{min\{\frac{N}{p},\frac{M}{p}\}}\mu(t)*\frac{N}{T}\frac{M}{T}\)。

• Enumerate what \ (T \) , are:

• \(\sum_{T=1}^{min\{N,M\}}\frac{N}{T}\frac{M}{T}\sum_{k|T,k\in primes}\mu(\frac{T}{k})\)。

• This latter may be pretreated prefix and a summing \ (O (1) \) query.

• Divisible by a preceding block may be optimized to \ (O (\ n-sqrt {}) \) .

• #include<bits/stdc++.h>
  using namespace std;
  const int maxn = 1e7 + 1e5;
  int mu[maxn], primes[maxn], cnt;
  long long sum[maxn], f[maxn];
  bool vis[maxn];
  void get_mu(int n)
  {
      mu[1] = 1;
      for(int i = 2; i <= n; i++)
      {
          if(!vis[i])
          {
              primes[++cnt] = i;
              mu[i] = -1;
          }
          for(int j = 1; primes[j] <= n/i; j++)
          {
              vis[primes[j]*i] = 1;
              if(i % primes[j] == 0) break;
              else mu[i*primes[j]] = -mu[i];
          }
      }
      for(int i = 1; i <= cnt; i++)
          for(int j = 1; primes[i] <= n/j; j++)
              f[j*primes[i]] += mu[j];
      for(int i = 1; i <= n; i++)
          sum[i] = sum[i-1] + f[i];
  }
  
  int main()
  {
      get_mu(10000000+5);
      int T; scanf("%d", &T);
      while(T--)
      {
          int n, m;
          long long ans = 0;
          scanf("%d%d", &n, &m);
          if(n > m) swap(n, m);
          for(int l = 1, r; l <= n; l = r+1)
          {
              r = min(n/(n/l), m/(m/l));
              ans += 1ll*(1ll*(n/l)*1ll*(m/l) * (sum[r]-sum[l-1]));
          }
          printf("%lld\n", ans);
      }
      return 0;
  }