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[323167] 主题： 一道程序题
作者： little (渺小・Happy^_^)
标题： 一道程序题
来自： 192.168.*.*
发贴时间： 2005年01月18日 12:49:22 (UTC +08:00)
长度： 2724字
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 <&g
t; 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 &l
t;> 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 * 10 + 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


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, result;
  int x, y, z;

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

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

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

if (pos == MAX_DIGIT)
{
//abc * d = efgh
x = result * 100 + result * 10 + result;
y = result;
z =
result * 1000 + result * 100 + result * 10 +
result;
if (x * y == z)
printf ("%d * %d = %dn", x, y, z);
//ab * cd = efgh
x = result * 10 + result;
y = result * 10 + result;
if ((x <= y) && (x * y == z))
printf ("%d * %d = %dn", x, y, z);

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

      digit] = 0;
      result = 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。

--
※作者已于 2005-01-18 18:28:10 修改本文※

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