题目：
    使用1-8这8个数字，组成abc*d=efgh或者ab*cd=efgh形式的等式。要求，每个数字被使
用1次。求出所有不重复的解。


解答一（VB）
    比较容易想到但十分繁琐的解题思路
Dim i, j, k, l, m, n, o, p As Integer
For i = 1 To 8
For j = 1 To 8
  Do While j = i And j < 8
  j = j + 1
 Loop
For k = 1 To 8
 Do While (k = j Or k = i) And k < 8
  k = k + 1
 Loop
For l = 1 To 8
 Do While (l = k Or l = j Or l = i) And l < 8
  l = l + 1
 Loop
For m = 1 To 8
 Do While (m = l Or m = k Or m = j Or m = i) And m < 8
  m = m + 1
 Loop
For n = 1 To 8
 Do While (n = m Or n = l Or n = k Or n = j Or n = i) And n < 8
  n = n + 1
 Loop
For o = 1 To 8
 Do While (o = n Or o = m Or o = l Or o = k Or o = j Or o = i) And o < 8
  o = o + 1
 Loop
For p = 1 To 8
 Do While (p = o Or p = n Or p = m Or p = l Or p = k Or p = j Or p = i) And p < 
8
  p = p + 1
 Loop
If p <> o And p <> n And p <> m And p <> l And p <> k And p <> j And p <> i And 
o <> n And o <> m And o <> l And o <> k And o <> j And o <> i And j <> i And k <
> j And k <> i And l <> k And l <> j And l <> i And m <> l And m <> k And m <> j
 And m <> i And n <> m And n <> l And n <> k And n <> j And n <> i Then
  If (i * 100 + j * 10 + k) * l = m * 1000 + n * 100 + o * 10 + p Then
   Print i; j; k; "*"; l; "="; m; n; o; p
  ElseIf (i * 10 + j) * (k * 10 + l) = m * 1000 + n * 100 + o * 10 + p And i * 1
0 + j < k * 10 + l Then
   Print i; j; "*"; k; l; "="; m; n; o; p
  End If
End If
Next p
Next o
Next n
Next m
Next l
Next k
Next j
Next i
[lml]

解答二（C）
    一种效率较高但比较难理解的解题思路
#include <stdio.h>
#define MAX_DIGIT 8
#define BUFFER_LEN 9

int
main ()
{
  int i, pos = 0, digit[BUFFER_LEN], result[BUFFER_LEN];
  int x, y, z;

  for (i = 0; i < BUFFER_LEN; i++)
    digit[i] = 0;

  for (i = 0; i < BUFFER_LEN; i++)
    result[i] = 0;

  while (pos >= 0)
    {
      //Search available digit
      for (i = result[pos] + 1; i <= MAX_DIGIT; i++)
    {
      if (digit[i] == 0)
        {
          digit[result[pos]] = 0;
          digit[i] = 1;
          result[pos++] = i;
          i = result[pos];

          if (pos == MAX_DIGIT)
        {
          //abc * d = efgh
          x = result[0] * 100 + result[1] * 10 + result[2];
          y = result[3];
          z =
            result[4] * 1000 + result[5] * 100 + result[6] * 10 +
            result[7];
          if (x * y == z)
            printf ("%d * %d = %dn", x, y, z);
          //ab * cd = efgh
          x = result[0] * 10 + result[1];
          y = result[2] * 10 + result[3];
          if ((x <= y) && (x * y == z))
            printf ("%d * %d = %dn", x, y, z);

          pos--;
        }
        }
    }
      if (pos == 0)
    break;

      digit[result[pos]] = 0;
      result[pos--] = 0;
    }

  return 0;
}

    个人推荐使用C语言实现的那种方法，不过实在不懂的用VB的那种也可以。

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※ 作者已于 2005-01-18 18:28:10 修改本文※