4 solutions

  • 11
    @ 2024-3-21 16:36:03
    /*
    观察每个对角线的起点和每次填写的次数,按照右上到左下一次填数即可 
    
    */
    #include<bits/stdc++.h>
    using namespace std;
    const int N=20;
    int g[N][N];
    int main()
    {
    	int n;
    	cin>>n;
    	int cnt=1;
    	for(int i=1;i<=n;i++) //填写第i列的起点 
    	{
    		for(int x=1,y=i;x<=i;x++,y--) //从(1,i) 
    		{
    			g[x][y]=cnt++;
    		}
    	}
    	for(int i=1;i<=n;i++)
    	{
    		for(int j=1;j<=n-i+1;j++)
    		{
    			cout<<g[i][j]<<" ";
    		}
    		cout<<endl;
    	}
    	return 0;
    }
    
    
    
    • 2
      @ 2026-1-11 9:22:32
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      1
      
      • 0
        @ 2026-8-12 21:21:10
        #include<bits/stdc++.h>
        using namespace std;
        int n;
        int main()
        {
        	cin>>n;
        	int t=1,d=1;
        	for(int i=1;i<=n;i++){
        		int r=t,e=d;
        		for(int j=1;j<=n-i+1;j++){
        			
        			
        			cout<<r<<" ";
        			r+=e;
        			e++;
        		}
        		d++;
        		t+=i+1;
        		cout<<endl;
        	}
        	return 0;
        }
        
        • -5
          @ 2024-6-12 20:06:07
          #include<bits/stdc++.h>
          using namespace std;
          const int N=11;
          int n[N][N];
          int main(){
          	int qwert;
          	cin>>qwert;
          	int asdf=1;
          	for(int i=1;i<=qwert;i++)
          	{
          		int l=i;
          		int p=1;
          		while(l>=1)
          		{
          			n[p][l]=asdf;
          			asdf++;
          			p++;
          			l--;			
          		}
          	}
          	for(int q=1;q<=qwert;q++)
          	{
          		for(int d=1;d<=qwert;d++)
          		{
          			if(n[q][d]==0)
          			{
          				continue;
          			}
          			cout<<n[q][d]<<" ";
          		}
          		cout<<endl;
          	}
          	return 0;
          }
          
          
          • @ 2026-8-12 21:23:28

            cxy675(专注于发布"正确"的题解) (黎锦宸)

        • 1

        Information

        ID
        81
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        5
        Tags
        (None)
        # Submissions
        88
        Accepted
        45
        Uploaded By