2022.02.14

Today I wrote an interesting topic , People are numb

Input ： The first line is an integer , Indicates the number of test cases . Every subsequent line , There is an integer n(1 < = n < = 30 ).

Output ： For each n×n Square array , Please refer to the output sample , Output coiled dragon array . To align , Each array element occupies 4 Character width , also , Insufficient 4 Press right to align the output of characters . There is a blank line after the output of each test case . Note that there are no spaces at the end of each line .

### Input examples

3
1
2
5

### Output example

   1

   1   2
   4   3

   1   2   3   4   5
  16  17  18  19   6
  15  24  25  20   7
  14  23  22  21   8
  13  12  11  10   9

This question is OK , Of course, the process is troublesome

#include<stdio.h>
#include<string.h>
int main()
{
   int n,m,i,j,k,a[100][100]={0};
   int num;
   scanf("%d",&n);
    while(n--)
   {
       num=1;
       scanf("%d",&m);
       for(j=0;j<=m/2;j++)
       {
           for(k=j;k<m-j;k++)
               a[j][k]=num++;
           for(k=j+1;k<m-j-1;k++)
            a[k][m-j-1]=num++;
           for(k=m-j;k>j;k--)
            a[m-1-j][k-1]=num++;
           for(k=m-j-2;k>j;k--)
            a[k][j]=num++;
       }
       if(m%2!=0)
       a[m/2][m/2]=m*m;
       for(i=0;i<m;i++)
       {
           for(k=0;k<m;k++)
            printf("%4d",a[i][k]);
            printf("\n");
       }
        printf("\n");
   }
    return 0;
}

  premium ：

Problem description

　　 To take a number from a loop is to take a number along the edge of a matrix , If the current direction of countless or have taken , Turn left 90 degree . It starts at the top left corner of the matrix , Direction down .

Input format

　　 The first line of input is no more than 200 The positive integer m, n, The rows and columns that represent the matrix . Next m Every line n It's an integer , To represent this matrix .

Output format

　　 There is only one line of output , common mn Number , The result of taking data for the input matrix . Numbers are separated by a space , Don't have extra space at the end of line .

The sample input

3 3
1 2 3
4 5 6
7 8 9

Sample output

1 4 7 8 9 6 3 2 5

The sample input

3 2
1 2
3 4
5 6

Sample output

1 3 5 6 4 2

At first glance, there is no difference between the two questions , But the feeling is different （ The child has fainted in the toilet ）

 

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,m,a[300][300],i,j;
    cin>>n>>m;
    int temp;
    if(n>=m)
        temp=m;
    else
        temp=n;
    for(i=0;i<n;i++)
    {
        for(j=0;j<m;j++)
            cin>>a[i][j];
    }
    if(n==1||m==1)
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(i==0&&j==0)
                    cout<<a[i][j];
                else
                    cout<<" "<<a[i][j];
            }
        }
        else
        {
    for(i=0;i<(temp+1)/2;i++)
    {
        if(i==(temp+1)/2-1&&temp==n&&temp%2!=0)
        {
            for(j=i;j<m-i;j++)
                cout<<" "<<a[n/2][j];
            break;
        }
        if(i==(temp+1)/2-1&&temp==m&&temp%2!=0)
        {
             for(j=i;j<n-i;j++)
                cout<<" "<<a[j][m/2];
            break;
        }
        for(j=i;j<n-i-1;j++)
        {
            if(i==0&&j==0)
                cout<<a[j][i];
            else
                cout<<" "<<a[j][i];
        }
        for(j=i;j<m-i-1;j++)
            cout<<" "<<a[n-1-i][j];
        for(j=n-1-i;j>i;j--)
            cout<<" "<<a[j][m-1-i];
        for(j=m-i-1;j>i;j--)
            cout<<" "<<a[i][j];
    }
        }
    return 0;
}

It's easy to get wrong ：1. To determine whether the horizontal and vertical coordinates of the array are 1, namely ：int a【1】【x】 or int a【x】【1】;

               2. Print back to determine the parity of the horizontal and vertical coordinates of the array ;

               3. The number of outermost cycles should be judged by the size of the horizontal and vertical coordinates ;

Two and a half hours just to get one question right , I can't write anymore .