请设计一个程序，用1—9这九个数字组成3个三位的平方数，要求每个数字只能使用一次。例如：361、529、784就是符合条件的解，因为它们本身既是一个平方数，而且1—9这九个数字在这三个平方数中中出现了一次。

CLS
DIM a(100), b(9)
FOR i = 10 TO 31
a(i) = i * i
NEXT i
FOR i = 1 TO 29
FOR j = i + 1 TO 30
   FOR k = j + 1 TO 31
    a$(1) = LTRIM$(STR$(a(i)))
    a$(2) = LTRIM$(STR$(a(j)))
    a$(3) = LTRIM$(STR$(a(k)))
    FOR a = 1 TO 3
     FOR b = 1 TO 3
      x$ = MID$(a$(a), i, 1)
      b(VAL(x$)) = b(VAL(x$)) + 1
      IF b(VAL(x$)) = 2 THEN GOTO 1
     NEXT b
    NEXT a
    PRINT a(i); a(j); a(k)
1   FOR a = 0 TO 9
     b(a) = 0
    NEXT a
   NEXT k
NEXT j
NEXT i
END


CLS
FOR i = INT(SQR(100)) TO INT(SQR(999))
j = i * i
j$ = STR$(j)
FOR k = 2 TO 4
a$(k - 1) = MID$(j$, k, 1)
NEXT k
IF a$(1) <> a$(2) AND a$(1) <> a$(3) AND a$(2) <> a$(3) THEN PRINT j;
NEXT i
END

完整程序：
CLS
h = INT(SQR(999)) - INT(SQR(100))
DIM m(h)
FOR i = INT(SQR(100)) TO INT(SQR(999))
j = i * i
j$ = STR$(j)
FOR k = 2 TO 4
a$(k - 1) = MID$(j$, k, 1)
NEXT k
IF a$(1) <> a$(2) AND a$(1) <> a$(3) AND a$(2) <> a$(3) THEN n = n + 1: m(n) = j
NEXT i
FOR x = 1 TO n - 2
FOR y = x + 1 TO n - 1
FOR z = y + 1 TO n
x$ = RIGHT$(STR$(m(x)), 3)
y$ = RIGHT$(STR$(m(y)), 3)
z$ = RIGHT$(STR$(m(z)), 3)
p$ = x$ + y$ + z$
FOR i = 1 TO 9
p$(i) = MID$(p$, i, 1)
NEXT i
g = 1
FOR i = 1 TO 8
FOR j = i + 1 TO 9
IF p$(i) = p$(j) THEN g = 0: EXIT FOR ELSE g = 1
NEXT j
IF g = 0 THEN EXIT FOR
NEXT i
IF g = 1 THEN PRINT m(x); m(y); m(z)
NEXT z, y, x
END

这些代码的共同特点,都堆在一起,可读性差,可修改性差.
只要历遍所有组合就可筛选出答案来