Spiral square PTA

So-called “ Spiral square ”, It refers to any given N, take 1 To N×N The number from the top left corner is 1 Start with a grid , Fill in in clockwise spiral direction N×N In the square of . This problem requires the construction of such a spiral square matrix .

Input format ：

Input gives a positive integer on a line N（<10）.

Output format ：

Output N×N Spiral matrix of . Each row N A digital , Each number accounts for 3 position .

sample input ：

5

sample output ：

  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

Compare the two ideas ：

What I think is to the right , Down , towards the left , Up to the end , Set flag subscript , Proceed to the next step .

#include <stdio.h>
int main(){
    int n,arr[100][100]={0};
    scanf("%d",&n);
    int i=0,j=-1,k=1,zuo=0,shang=1;
    while(k<=n*n&&n!=1){
        while(1){
            if(arr[i][j]==n*n){break;}
            j++;
            arr[i][j]=k++;
            if(j==n-i-1){
                break;
            }
        }
        while(1){
            if(arr[i][j]==n*n){break;}
            i++;
            arr[i][j]=k++;
            if(i==j){
                break;
            }
        }
        while(1){
            if(arr[i][j]==n*n){break;}
            j--;
            arr[i][j]=k++;
            if(j==zuo){
                zuo++;
                break;
            }
        }
        while(1){
            if(arr[i][j]==n*n){break;}
            i--;
            arr[i][j]=k++;
            if(i==shang){
                shang++;
                break;
            }
        }
    }
    if(n==1){arr[0][0]=1;}
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            printf("%3d",arr[i][j]);
        }
        printf("\n");
    }
    return 0;
}

Learn the thought of the great God , Upper left and lower right n-1 individual , Then operate ;

#include<stdio.h>
int main()
{
	int n;
	scanf("%d",&n);
	int arr[n][n],i,j,k=1,c=n-1,x=0,y=0;
	for(i=0;i<n/2;i++)
	{
		for(j=0;j<c;j++) arr[x][y++] = k++;
		for(j=0;j<c;j++) arr[x++][y] = k++;
		for(j=0;j<c;j++) arr[x][y--] = k++;
		for(j=0;j<c;j++) arr[x--][y] = k++;
		x++;y++;c-=2;
	}
	if(n%2) arr[n/2][n/2] = k;
	for(i=0;i<n;i++)
	{
		for(j=0;j<n;j++)
	   {
		     printf("%3d",arr[i][j]);
	   }
	   printf("\n");
	}
 }