#define begin   {
#define end     }
#define then
#define repeat   do{
#define until(x)   }while(not(x))
#define procedure void
#define mod %
#define and &&
#define not !
#define div /
#define MAXorder 13
/*** Magic SQUARE(魔方陣)也是暸解array很好的例子 ***/
/*** 魔方陣是指橫的直的斜的加起來都一樣的連續數字方陣***/
/* For example, the following is a Magic Square of order 4:
       1  15  14   4  
      12   6   7   9
       8  10  11   5
      13   3   2  16
************************** 偉大的數學家們發現:
        四階魔方陣共有880個, 上例只是其中之一
    不過三階魔方陣則只有一個!
********************************/
/***** Order 為奇數的Magic Square 可用一簡單方法產生:
      在C++課本中有, 第一個說出在第幾頁並
      且把原文整題照key in 並follow up 於此的可加期中考3分
************************************************/
int x[1+MAXorder] [1+MAXorder] ;   /*這是 global array , 整個程式都有效*/
procedure outx(int n);
procedure oddmagic(int n);
int main()
begin
    int i,j,k;
    int n; 
    char yesno;
  repeat
     printf("Generate Magic Square, order=?");
     scanf("%d", &n);
     while(not (( (n>=3) and (n <=MAXorder)) and (n mod 2 == 1)) )  begin
        /*  n 要在範圍內且要奇數我們才做 */
        if(not ((n>=3) and (n<= MAXorder)) )then begin
           printf(" order must be in 3..%d", MAXorder);
           if(n mod 2 == 0) then printf(", and must be an odd number");
        end else begin
           if(n mod 2 == 0) then printf("Order should be an odd number");
        end;
        printf(", order=?");
        scanf("%d", &n);
     end;
     oddmagic(n);    /* call the procedure oddmagic() */
                 /* note that the array x is global */
     outx(n);      /* print out the square x of order n */
     printf("One more again(N,Y)?"); scanf(" %c", &yesno);
  until( (yesno !='y') and (yesno!='Y'));
end;
procedure oddmagic(int n)
 begin
     int row,col,value;
     int t1,t2;
    value=0;
    row= 1-1;            /* 第一列的上一列 */
    col= (n+1) div 2;    /* 中間那格 */
    for(t1=1; t1<=n; t1++) begin
       /* 到下一列 */
       row=row+1; if(row > n)then row=1;
       value=value+1; x[row][col]=value; /* 填下一值 */
       for(t2=1; t2<= n-1; t2++) begin
          /* 到右上角一格 */
          row=row-1; if(row < 1)then row=n;
          col=col+1; if(col > n)then col=1;
          value=value+1; x[row][col]=value;
       end;
    end;
 end;
procedure outx(int n)
 begin
   int  i,j;
   for(i=1; i<=n; i++) begin
       for(j=1; j<=n; j++) begin
          printf("%5d", x[i][j]);
       end;
       printf("\n");;
   end;/* for i */
 end; /*** outx ***/
/*******************************************
From: cclo@csie.nctu.edu.tw (想加 RAM 的小孩)
    哇咧....這麼好賺！不 key 一下對不起自己....
    (page 806)
    The following is a procedure for constructing an n*n magic square for
  any odd integer n. Place 1 in the middle of the top row. Then after in-
  teger k has been placed, move up one row and one column to the right to 
  place the next integer k+1, unless one of the followig occurs:
   i) If a move takes you above the top row in the jth column, move to the
      bottom of the jth column and place the integer k+1 there.
  ii) If a move takes you outside to the right of the square in the ith 
      row, place k+1 in the ith row at the left side.
 iii) If a move takes you to an already filled square or if you move out 
      of the square at the upper right-hand corner, place k+1 immediately
      below k.
/**********************************************************************/