3 solutions

  • 2
    @ 2024-8-20 15:15:55
    #include<bits/stdc++.h>
    using namespace std;
    const int N=200010;
    vector<int> g[N];
    int main()
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n/i;j++)
            {
                g[i*j].push_back(i); //将i放到i的倍数后面 
            }
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d:",i);
            for(int j=0;j<g[i].size();j++)
            {
                printf("%d ",g[i][j]);
            }
            printf("\n");
        }
        return 0;
    }
    
    • 1
      @ 2025-3-15 15:12:15
      #include<bits/stdc++.h>
      using namespace std;
      const int N=200010;
      vector<int> g[N];
      int main()
      {
          int n;
          scanf("%d",&n);
          for(int i=1;i<=n;i++)
          {
              for(int j=1;j<=n/i;j++)
              {
                  g[i*j].push_back(i); //将i放到i的倍数后面 
              }
          }
          for(int i=1;i<=n;i++)
          {
              printf("%d:",i);
              for(int j=0;j<g[i].size();j++)
              {
                  printf("%d ",g[i][j]);
              }
              printf("\n");
          }
          return 0;
      }
      • -2
        @ 2024-7-6 14:41:43
        #include<bits/stdc++.h>
        using namespace std;
        vector<int> g;
        int main()
        {
        	long long n;
        	for(int i=0;i<n;i++)
        	{
        		for(int j=1;j<=n*i;j++)
        		{
        			g[j&i].push_back(j);
        		}
        	}
        	for(int i=1;i<=n*j;i++)
        	{
        		for(int j=1;j<g[i].size()*i;j++)
        		{
        			printf("%d ",g[i][j]);
        		}
        		printf("\n"); 
        	}
        	return 0;
        }
        
        
        • 1

        Information

        ID
        182
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        3
        Tags
        # Submissions
        89
        Accepted
        15
        Uploaded By