簡單的說,就是一個從左上由1依序開始填,但遇對角線不填,另一個由左上由16開始填,但只填在對角線,再將兩個合起來就是解答了;如果N大於2,則以 4X4為單位畫對角線:
至於對角線的位置該如何判斷,有兩個公式,有興趣的可以畫圖印證看看,如下所示:
左上至右下:j % 4 == i % 4
右上至左下:(j % 4 + i % 4) == 1
實作
#include <stdio.h>
#include <stdlib.h>
#define N 8
int main(void) {
int i, j;
int square[N+1][N+1] = {0};
for(j = 1; j <= N; j++) {
for(i = 1; i <= N; i++){
if(j % 4 == i % 4 || (j % 4 + i % 4) == 1)
square[i][j] = (N+1-i) * N -j + 1;
else
square[i][j] = (i - 1) * N + j;
}
}
for(i = 1; i <= N; i++) {
for(j = 1; j <= N; j++)
printf("%2d ", square[i][j]);
printf("\n");
}
return 0;
}
public class Matrix {
public static int[][] magicFourN(int n) {
int[][] square = new int[n+1][n+1];
for(int j = 1; j <= n; j++) {
for(int i = 1; i <= n; i++){
if(j % 4 == i % 4 || (j % 4 + i % 4) == 1)
square[i][j] = (n+1-i) * n -j + 1;
else
square[i][j] = (i - 1) * n + j;
}
}
int[][] matrix = new int[n][n];
for(int k = 0; k < matrix.length; k++) {
for(int l = 0; l < matrix[0].length; l++) {
matrix[k][l] = square[k+1][l+1];
}
}
return matrix;
}
public static void main(String[] args) {
int[][] magic = Matrix.magicFourN(8);
for(int k = 0; k < magic.length; k++) {
for(int l = 0; l < magic[0].length; l++) {
System.out.print(magic[k][l] + " ");
}
System.out.println();
}
}
}