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来自「以电子书的形式收集了VB一些常见问题解决方法,可以很方便的查找自己需要解决的问题」· HTML 代码 · 共 23 行
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<title>Re: 交换法</title>
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<h1>Re: 交换法</h1>
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Posted by <a href="mailto:knight@ms1.rpticomp.com.tw">Knight</a> on November 10, 1998 at 10:59:14:<p>
In Reply to: <a href="6812.html">VB有无 "产生不重复乱数" 之功能函数?</a> posted by 小Sam on October 28, 1998 at 10:24:22:<p>
: 请问各位传统Basic老手或新新人类VB使用者,<br>: VB有无 "产生不重复乱数" 之功能函数?<br>: (意指不用任何技巧 ,而是一个函数所提供的功能?)<br>: 若有 ,是否会因乱数范围大小而影响产生速度?<br>: 如:1~10000000中取10000组不重复的数字 ,<br>: 取到后来会不会越取越慢?<br>: 若无 ,<br>: 有无较简单方法而又不会造成越取越慢?<p>: Thanks...小Sam<br>以下是以交换法达成不重覆乱数<br>请大家指教<br>10000000 组数字在p-133约5 min<br>Option Explicit<br>Const Random_Num = 10000000<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<p>
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