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    <title>Re: 简单问题: 如果有20个数字,1 To 20,如何可以Random地出现不同次序?</title>
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      <h1>Re: 简单问题: 如果有20个数字,1 To 20,如何可以Random地出现不同次序?</h1>
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Posted by <a href="mailto:knight@ms1.rpticomp.com.tw">knight</a> on November 16, 1998 at 09:30:35:<p>
In Reply to: <a href="7314.html">Re: 简单问题: 如果有20个数字,1 To 20,如何可以Random地出现不同次序?</a> posted by 匡 on November 12, 1998 at 17:07:31:<p>
: : 既然你说是简单问题,就再想一想吧! 把你的方法 post 出来让大家参考。<br>: : 我是有一个办法,不过下周再告诉你。<p>: 如困是7个数字,最简单的方法....,但要排列20个数字就麻烦了,怎么办?<p>: Do<br>: Randomize<br>: am% = Int(7 * Rnd)<p>: bm% = Int(7 * Rnd)<p>: cm% = Int(7 * Rnd)<p>: dm% = Int(7 * Rnd)<p>: em% = Int(7 * Rnd)<p>: Loop Until am% <> bm% And am% <> cm% And am% <> dm% And am% <> em% And bm% <> cm% And bm% <> dm% And bm% <> em% And cm% <> dm% And cm% <> em% And dm% <> em%<br>以下是以交换法达成不重覆乱数<br>Option Explicit<br>Const Random_Num = 20<br>Dim NUM(Random_Num) As Long<br>End Sub<br>Private Sub Form_Load()<br>Dim Change_Num1&, Change_Num2&, Change_Buf&, Count_Num&<br>Randomize<br>For Count_Num = 0 To Random_Num<br>NUM(Count_Num) = Count_Num<br>Next<br>For Count_Num = 0 To Random_Num<br>Change_Num1 = Int(Rnd * Random_Num) + 1<br>Change_Num2 = Int(Rnd * Random_Num) + 1<br>Change_Buf = NUM(Change_Num1)<br>NUM(Change_Num1) = NUM(Change_Num2)<br>NUM(Change_Num2) = Change_Buf<br>Next<br>End Sub<br>先在阵列中放入0-20<br>再以乱数交换任意两数<br>
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