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
      @ 2025-7-21 10:47:33
      #include<bits/stdc++.h>
      using namespace std;
      const int N=200010;
      vector<int> g[N];
      int main()
      {
      	int n;
      	cin>>n;
      	for(int i=1;i<=n;i++)
      	{
      		for(int j=1;j<=n/i;j++)
      		{
      			g[i*j].push_back(i);
      		}
      	}
      	for(int i=1;i<=n;i++)
      	{
      	cout<<i<<":";
      		for(int j=0;j<g[i].size();j++)
      		{
      			cout<<g[i][j]<<" ";
      		}
      		cout<<endl;
      	}
      	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
        88
        Accepted
        14
        Uploaded By