HLG 1564 spiral matrix (fun C language)

 

 

Description
For a given number n, You need to print the spiral matrix of n * n.

For example, if n = 3, the output is:

1 2 3
8 9 4
7 6 5

Input

Multiple groups of test data, each of which contains an integer n (1 <=n <= 32)

Output

For each group of test data, an n * n spiral matrix is output, which is defined in the topic description.

In a set of test data, the character width of each number is the maximum digit plus 1 in the set of data, for example, the spiral matrix of 3*3. The maximum value is 9, then each number occupies 2 Characters in width.

Two groups of test data are separated by one empty row.

Sample Input
1
2
3
Sample Output
1

1 2
4 3

1 2 3
8 9 4
7 6 5

 

The Code is as follows ~~~)

 

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #define MAXN 105#define RST(N)memset(N, 0, sizeof(N))using namespace std;int Kesha[105][105], cas = 0, n;int main(){    while(~scanf(%d, &n)) {        if(cas++) printf();        int x, y, ans = 0;        RST(Kesha);        ans = Kesha[x=0][y=0] = 1;        while(ans < n * n) {            while(y + 1 < n && !Kesha[x][y+1]) Kesha[x][++y] = ++ans;            while(x + 1 < n && !Kesha[x+1][y]) Kesha[++x][y] = ++ans;            while(y - 1 >= 0 && !Kesha[x][y-1]) Kesha[x][--y] = ++ans;            while(x - 1 >= 0 && !Kesha[x-1][y]) Kesha[--x][y] = ++ans;        }        int t = n * n;        for(int i=0; i
     
      =10 && t<=99) printf(%3d, Kesha[i][j]);                else if(t>=100 && t<=999) printf(%4d, Kesha[i][j]);                else if(t >= 1000 ) printf(%5d, Kesha[i][j]);            }            printf();        }    }    return 0;}