C language-Print magic squares (each row, each column, the sum of the diagonal equals)

"One" Magic square Description:

Rubik's Cube array is a n*n matrix;

Each row of the matrix, each column, the sum of the diagonal are equal;

Example of the second magic Phalanx:

Third Order Magic Phalanx:

8 1 6

3 5 7

4 9 2

The sum of each line: 8+1+6=15;

3+5+7=15;

4+9+2=15;

The sum of each column: 8+3+4=15;

1+5+9=15;

6+7+2=15;

Diagonal sum: 8+5+2=15;

6+5+4=15;

"Three" magic square calculation Law (row, column starting at 1):

1. Place "1" in the first row, one column in the middle;

2. From 2 onwards to n*n the following rules:

Each number of rows is reduced by 1 compared to the previous number of rows;

Each number of columns is 1 more than the previous column;

3. When a number behaves 1, the next number behaves n;

4. When a sequence number is n, the next number of columns is 1, and the number of rows is reduced by 1;

5. If the position determined by the above rules has a number, or the previous digit 1th row n,

The next digit position is directly below the previous number (the number of rows minus 1, the number of columns is unchanged);

"Four" source code:

1 #define_crt_secure_no_warnings2#include <stdio.h>3#include <stdlib.h>4#include <math.h>5 6 #defineN 57 8 intMain ()9 {Ten     intA[n][n] = {0}; One     intCount =1; A     introw =0, cul = N/2; -      while(Count <= NN) -     { theA[row][cul] =count; -         inti =Row; -         intj =Cul; -         if(i = =0) +         { -i = N-1; +         } A         Else at         { -i = i-1; -         } -j = (J +1) %N; -         if(a[i][j]!=0|| (row==0&&cul==n-1)) -         { ini = row +1; -j =Cul; to         } +row =i; -Cul =J; thecount++; *     } $ Panax Notoginseng      for(inti =0; i < N; i++) -     { the          for(intj =0; J < N; J + +) +         { Aprintf"%3d", A[i][j]); the         } +printf"\ n"); -     } $  $System"Pause"); -}

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C language-Print magic squares (each row, each column, the sum of the diagonal equals)