最近查看了一些魔方阵的资料。网上有很多生成奇数阶魔方阵的方法， 也有一些生成双偶数阶魔方阵的方法。回去重点思考一下单偶数阶魔方阵的生成方法。今天终于编写调试（在Turboc 2.0中）成功，现在与大家分享。
有关魔方阵的资料，可参考https://www.bc-cn.net/bbs/dispbbs.asp?boardID=5&ID=28099

#include <stdio.h>
main()
{
int i,j,n,k,p,q,t,sum,a[20][20]={0};
printf("Please input N:\n");
scanf("%d",&n);
sum=n*(n*n+1)/2;
t=n;
if(n<3||n>20)
printf("The number is out of range.");
else if(n%2!=0) /* According to the De La Loubere's Method for odd level Rubik Cube */
{
k=1;
i=0;
j=(n-1)/2;
while(k<=n*n)
{
while(a[i][j]==0)
{
a[i][j]=k;
k++;
if(i==0)
i=n-1;
else i=i-1;
if(j==n-1)
j=0;
else j=j+1;
}
if(i==n-1)
i=1;
else i=i+2;
if(j==0)
j=n-1;
else j=j-1;
}
}
else {
if(n%4!=0) /* According to the Yanghui's Extended Method for quadruple level Rubik Cube */
n-=2;
p=2;
q=n*n+1;
for(i=1;i<=n;)
{
while(i%4==0||i%4==1)
{
if(i%4==0)
{
p+=2;
q+=1;
}
else q-=1;
a[i][1]=q;
q-=3;
for(j=3;j<=n;j+=4)
{
a[i][j-1]=p;
a[i][j]=p+1;
p+=4;
}
for(k=4;k<n;k+=4)
{
a[i][k]=q;
a[i][k+1]=q-1;
q-=4;
}
a[i][k]=q;
i+=1;
}
while(i%4==2||i%4==3)
{
if(i%4==2)
{
p-=1;
q-=2;
}
else p+=1;
a[i][1]=p;
p+=3;
for(j=4;j<n;j+=4)
{
a[i][j]=p;
a[i][j+1]=p+1;
p+=4;
}
for(k=3;k<=n;k+=4)
{
a[i][k-1]=q;
a[i][k]=q-1;
q-=4;
}
a[i][j]=p;
i+=1;
}
}
if(t%4!=0) /* According to the Expand Level Method for single even level Rubik Cube */
{
n+=2;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
a[i][j]+=2*n-2;
a[0][0]=n*n-n;
a[n-1][n-1]=n+1;
a[n-1][0]=n;
a[0][n-1]=n*n-n+1;
for(k=1;k<=(n/2-1);k++)
{
a[2*k-1][0]=2*k;
a[2*k][0]=n*n-2*k;
a[0][2*k-2]=n*n-n-2*k+2;
a[0][2*k-1]=n+2*k;
}
a[1][0]=n*n-1;
a[n/2][0]=1;
a[n/2+1][0]=n/2+2;
a[0][n-2]=n/2+1;
for(k=1;k<n-1;k++)
{
a[n-1][k]=n*n+1-a[0][k];
a[k][n-1]=n*n+1-a[k][0];
}
}
}
if(n>=3&&n<=20)
{
if(n%4!=0)
{
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
printf("%3d ",a[i][j]);
putchar('\n');
}
}
else {
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%3d ",a[i][j]);
putchar('\n');
}
}
printf("The sum of the numbers in every row, column and diagonal is %d.\n",sum);
}
}

/* De La Loubere's Method: Place 1 in the first row, in the center of the column. Then go right and up one form,

place N+1, and continue go, contunue place N+1, if the form you go has already had a number, return back, go

down one form, place N+1, and contunue go right and up one form, place N+1, at last you will get odd level Rubik

Cube.

* * 1 * * * * 1 * * * * 1 * * * * 1 * * * * 1 8 * 17 24 1 8 15
* * * * * * * * * * * 5 * * * * 5 * * * * 5 7 * * 23 5 7 14 16
* * * * * * * * * * 4 * * * * 4 6 * * * 4 6 * * * 4 6 13 20 22
* * * * * * * * * * * * * * 3 * * * * 3 10 * * * 3 10 12 19 21 3
* * * * * * * * 2 * * * * 2 * * * * 2 * 11 * * 2 9 11 18 25 2 9 */

/* Yanghui's Extended Method: Divide the N*N phalanx into several 4*4 phalanxes. Mark the forms in every 4*4

phalanx as below. Place i in the i's form of * form, place N*N-i in the i's form of @ form just as below.

@ * * @ @ * * @ 01 02 03 04 05 06 07 08 64 02 03 61 60 06 07 57
* @ @ * * @ @ * 09 10 11 12 13 14 15 16 09 55 54 12 13 51 50 16
* @ @ * * @ @ * 17 18 19 20 21 22 23 24 17 47 46 20 21 43 42 24
@ * * @ @ * * @ 25 26 27 28 29 30 31 32 40 26 27 37 36 30 31 33
@ * * @ @ * * @ 33 34 35 36 37 38 39 40 32 34 35 29 28 38 39 25
* @ @ * * @ @ * 41 42 43 44 45 46 47 48 41 23 22 44 45 19 18 48
* @ @ * * @ @ * 49 50 51 52 53 54 55 56 49 15 14 52 53 11 10 56
@ * * @ @ * * @ 57 58 59 60 61 62 63 64 08 58 59 05 04 62 63 01 */

/* Expand Level Method: Firstly construct a N-2 Rubic Cube which is a quadruple level Rubik Cube (the area

of the *). Then add 2N-2 to all the numbers in the phalanx. Place N*N-N in the left top @ form, N+1 right bottom

@ form, N left bottom @ form,N*N-N+1 right top @ form. Divide fisrt column(except two ends of @ form) into 2

groups, - form is the first group, + form is another. Assign a[2k-1][0]=2k which is the first group,

a[2k][0]=N*N-2k which is the second.(k=1,2...2/N-1) Make somme adjustment as follows: a[1][0]=N*N-1, [n/2][0]=1,

a[n/2+1][0]=N/2+1. Otherwise, the sum of the first column will not be equal to N*(N*N+1)/2. Divide first row

(except back end of @ form) into 2 groups, left top @ and - form is the first group,+ form is another. Assign

a[0][2k-2]=N*N-N-2k+2 which is the first group, a[0][2k-1]=N+2k which is the second.(k=1,2...2/N-1)

Then # form can be caculated by the formula. For example, # form in the first row is equal to 2/N+1.

@ + - + # @ @ + - + # @ 30 + - + # 31 30 08 28 10 # 31 30 08 28 10 04 31
- * * * * # - 16 02 03 13 # - 26 12 13 23 # 35 26 12 13 23 # 35 26 12 13 23 02
+ * * * * # + 05 11 10 08 # + 15 21 20 18 # 34 15 21 20 18 # 34 15 21 20 18 03
- * * * * # - 09 07 06 12 # - 19 17 16 22 # 01 19 17 16 22 # 01 19 17 16 22 36
+ * * * * # + 04 14 15 01 # + 14 24 25 11 # 05 14 24 25 11 # 05 14 24 25 11 32
@ # # # # @ @ # # # # @ 06 # # # # 07 06 # # # # # 06 29 09 27 33 07 */

[此贴子已经被作者于2005-11-25 13:05:50编辑过]