只做了一半[em12][em12][em12]
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   2   5   9   12  20
   6   8   13  19
   7   14  18
   15  17
   16

DECLARE SUB printf ()
DIM SHARED num, x, dx, y, dy
CLS : INPUT "n=", n
CLS : n = 2 * (n - 1)
num = 0: x = 1: y = 1
dx = 0: dy = 0: printf
FOR i = 1 TO n STEP 2
   dx = 0: dy = 1: printf
   FOR j = 1 TO i
     dx = 1: dy = -1: printf
   NEXT j
   dx = 1: dy = 0: printf
   FOR j = 1 TO i + 1
     dx = -1: dy = 1: printf
   NEXT j
NEXT i

SUB printf
num = num + 1
x = x + dx: y = y + dy
LOCATE 2 * y, 4 * x: PRINT LTRIM$(STR$(num))
END SUB

djisfhckusdncasdghicuhskjc nauoidhrewyrqfgojdfcxnvjkcxhvioayriogquteroijhsldnvlchoiawurogiquwiogwfjglksmnvckljfsdhidvjehwurhguiohdsjgdJAXIOHSOIFHISADCLKSLSJVIJRIWGUAIWJGKLSCNVLKCSVKJLKSFJKLSJGKLFSJGSDGJUIOEWRUOIUOIJOJVLKMXCVLKC[em1]

[img]http://www.programfan.com/club/face/63.gif[/img]你什么意思啊？ 

终于解决了：
CLS
INPUT "n="; n
DIM a(n, n)
x = 1: y = 1: dx1 = 0: dy1 = 1: dx2 = 1: dy2 = -1
a(1, 1) = 1: a(n, n) = n ^ 2
FOR i = 2 TO n
  a(x + dx1, y + dy1) = a(x, y) + 1
  x = x + dx1: y = y + dy1
  a(n + 1 - x, n + 1 - y) = n ^ 2 + 1 - a(x, y)

  FOR j = 1 TO i - 1
    a(x + dx2, y + dy2) = a(x, y) + 1
    x = x + dx2: y = y + dy2
    a(n + 1 - x, n + 1 - y) = n ^ 2 + 1 - a(x, y)
  NEXT j

  SWAP dx1, dy1: SWAP dx2, dy2
NEXT i

CLS
FOR i = 1 TO n
FOR j = 1 TO n
  LOCATE n + 1 - i, 3 * j: PRINT a(j, i);
NEXT j
NEXT i

这是我编的程序

INPUT b:CLS:FOR a=1TO (b+1)\2:FOR c=0TO b-2*a+1:d=4*(a-1)*(b+1-a)+c+1:e=4*a-3:f=b-a+1:g=b-2*a+1:LOCATE a+c,e:?d;:LOCATE f,e+4*c:?d+f-a;:LOCATE f-c,4*f-3:?d-1+2*b-e;:LOCATE a,e+4*(g-c):?d+3*g;:LOCATE a,e:?d-c;:NEXT:NEXT

n需要>=1 <=19
共218字节

我和同屋一同学比赛编这道题，我用quickbasic4.5，他用tc2.0，方法任意，看谁的程序最短
呵呵不好意思，为了追求短小，一点可读性都没有了

我想，还是处理一下的好

CLS
INPUT "n=",n
CLS
FOR a=1 TO (b+1)\2
  FOR c=0 TO b-2*a+1
    LOCATE a+c,4*a-3
      PRINT 4*(a-1)*(b+1-a)+c+1;
    LOCATE b-a+1,4*a-3+4*c
      PRINT 4*(a-1)*(b+1-a)+b-2*a+2+c;
    LOCATE b-a+1-c,4*(b-a)+1
      PRINT 4*(a-1)*(b+1-a)+2*b-4*a+3+c;
    LOCATE a,4*a-3+4*(b-2*a+1)-4*c
      PRINT 4*(a-1)*(b+1-a)+1+3*(b-2*a+1)+c;
    LOCATE a,4*a-3
      PRINT 4*(a-1)*(b+1-a)+1;
  NEXT
NEXT

最后一个locate/print语句从计算方法上看，是完全多余的
但是经过几次试验之后发现删去之后，无法正确打印
肯定是我哪里想错了
谁能帮帮我啊

奇怪...竟然看错题了
不好意思，嘿嘿

对于这个，我总是建议：直接打！

CLS
我做出来了：
INPUT N
DIM A(N, N)
I = N: J = 1
S = (1 + N) * N / 2
FOR K = 1 TO S
    A(I, J) = K
    A(N + 1 - I, N + 1 - J) = N * N + 1 - K
    IF (I + J) MOD 2 = 0 THEN
       IF J = 1 THEN I = I - 1 ELSE I = I - 1: J = J - 1
    ELSE
       IF I = N THEN J = J + 1 ELSE J = J + 1: I = I + 1
    END IF
NEXT K
FOR I = 1 TO N
  FOR J = 1 TO N
      PRINT USING "####"; A(I, J);
  NEXT J
  PRINT
NEXT I

input "h，l:",h,l
dim s(h,l)
xx=1
yy=-1
for i=1 to h*l
  x=x+xx
  y=y+yy
  if x<1 or x>h or y<1 or y>l then swap xx,yy
  if x>h then 
     x=h
     y=y+2
  endif
  if y>l then 
     y=l
     x=x+2
  endif
  if x<1 then x=1
  if y<1 then y=1
  s(x,y)=i
next

for i=1 to h
  print
  for j=1 to l
    print using "###";s(i,j);
next j,i