1.8.22

22: Magic Square

Total time limit:

1000ms

Memory Limit:

65535kB

Describe

Magic Square is a very magical n*n matrix, it's each row, each column and the diagonal, plus the number is the same.
We can construct a magic square by the following methods. (Number of orders is odd)
1. The first digit is written in the middle of the first line
2. The next number is written in the upper right of the previous number:
A. If the number is in the first row, the next number is written in the last row, and the number of columns is the right column of the number
B. If the number is in the last column, the next number is written in the first column and the row number is the previous row of the number
C. If the number is in the upper-right corner, or there is a number at the top right of the number, the next digit is written below the number

Input

a number N (n<=20)

Output

The

magic square of 2n-1 * 2n-1 constructed

by the method

Sample input

3

Sample output

17 24 1 8 1523 5 7 14 164 6 13 20 2210 12 19 21 311 18 25 2 9

1 /*December 4, 2016 Openjudge daily water problem2 ———— 1.8.22 by Lxzy_zby*/3#include <cstdio>4#include <cstring>5 using namespacestd;6 BOOLb[101][101];7 inta[101][101];8 intMain ()9 {Ten     intn,n=2;//N defines the first assignment Onescanf"%d",&N); A     intx=0, y= (2*N)/2-1;//defining (x, y) initial coordinates -a[0][(2*N)/2-1]=1;//initializes the first row with an intermediate number of 1 -memset (b,0,sizeof(b));//BOOL Array Initialization theb[0][(2*N)/2-1]=1;//Initialize first row middle number bool value is 1 -     intAns= (2*n-1)*(2*n-1)+1, ans1=2*n-2; -      while(1) -     { +         if(n==1) Break;//processing n 1 of data separately -         if(x==0&AMP;&AMP;Y!=ANS1)//a case of +         { Ax=2*n-2, y+=1; ata[x][y]=N; -n++; -b[x][y]=1; -         } -         Else -         { in             if(y==ans1&&x!=0)//B Case -             { tox-=1, y=0; +a[x][y]=N; -n++; theb[x][y]=1;  *             } $             ElsePanax Notoginseng             { -                 if(b[x-1][y+1]==1|| x==0&&y==ans1&&b[0][ans1]==1)//C Case the                 { +X + +; Aa[x][y]=N; then++; +b[x][y]=1; -                 } $                 Else $                 { -                     if(b[x-1][y+1]==0)//Normal situation -                     { thex--; -y++;Wuyia[x][y]=N; theb[x][y]=1; -n++; Wu                     } -                 } About                  $             } -         } -         if(n==ans) -          Break;//if the last assignment has a coordinate value of (2n-1) * (2n-1) After the end of the loop, exit the loop A     } +      for(intI=0; i<=ans1;i++) the      { -           for(intj=0; j<=ans1;j++) $printf"%d", A[i][j]); theprintf"\ n"); the      } the     return 0; the}

1.8.22