2 solutions

  • 1
    @ 2024-6-1 15:39:09
    #include<bits/stdc++.h>
    using namespace std;
    const int N=1e6+10;
    int a[N],cnt[N],s[N];
    int main()
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            cnt[a[i]]++; //统计a[i]出现的次数 
        }
        for(int i=1;i<N;i++)
        {
            for(int j=i;j<N;j+=i) //将i出现的次数累加到i的倍数j中 
            {
                s[j]+=cnt[i];
            }
        }
        for(int i=1;i<=n;i++)
        {
            cout<<s[a[i]]-1<<endl;
        }
        return 0;
    }
    
    • 0
      @ 2024-7-10 14:51:28
      #include<bits/stdc++.h>
      using namespace std;
      const int N=1e6+10;
      int a[N],cnt[N],s[N];
      int main()
      {
         int n;
         cin>>n;
         for(int i=1;i<=n;i++)
         {
            cin>>a[i];
            cnt[a[i]]++;
         }
         for(int i=1;i<N;i++)
         {
            for(int j=1;j<=N/i;j++)
            {
               s[j*i]+=cnt[i];
            }
         }
         for(int i=1;i<=n;i++)
         {
            cout<<s[a[i]]-1<<endl;
         }
         return 0;
      }
      
      • 1

      Information

      ID
      971
      Time
      1000ms
      Memory
      256MiB
      Difficulty
      5
      Tags
      (None)
      # Submissions
      64
      Accepted
      14
      Uploaded By