关于解多元一次方程递归算法的引申问题
递归算法的运用(代码为网上转载)
Private Sub Command2_Click() '演示求X1+X2+X3+X4+X5=10非负整数解
Text1.Text = ""
Dim answer As String
answer = GETRESULT(5, 10, False) '赋值
Dim temp
temp = Split(answer, vbCrLf)
For i = 0 To UBound(temp)
temp(i) = "解" & i + 1 & ":" & vbTab & temp(i) 'add index
Next
answer = Join(temp, vbCrLf)
Text1.Text = "方程 X1+X2+X3+X4+X5=10 共有 " & UBound(Split(answer, vbCrLf)) + 1 & " 个非零整数解:" & v bCrLf & answer 'show all answer in textbox
End Sub
'求解函数
Function GETRESULT(ByVal n As Integer, ByVal SUM As Integer, Optional allowzero As Boolean = True) As String
Dim temp() As String, i As Long
If n = 2 Then '二元方程
If allowzero = True Then
ReDim temp(SUM)
For i = 1 To SUM ' allow zero
temp(i) = "X1=" & i & ",X2=" & SUM - i
Next
GETRESULT = Join(temp, vbCrLf)
Erase temp
Else
ReDim temp(1 To SUM - 1) 'forbid zero
For i = 1 To SUM - 1
temp(i) = "X1=" & i & ",X2=" & SUM - i
Next
GETRESULT = Join(temp, vbCrLf)
Erase temp
End If
End If
If n > 2 Then
If allowzero = True Then
ReDim temp(SUM)
For i = SUM To 0 Step -1 ' allow zero
temp(i) = Replace(GETRESULT(n - 1, i, True), vbCrLf, ",X" & n & "=" & SUM - i & vbCrLf) & ",X" & n & "=" & SUM - i
Next
GETRESULT = Join(temp, vbCrLf)
Erase temp
Else
If SUM < n Then MsgBox "无解!": Exit Function '无解情况
ReDim temp(1 To SUM - n + 1) 'not allow zero
For i = 1 To SUM - n + 1
temp(i) = Replace(GETRESULT(n - 1, SUM - i, False), vbCrLf, ",X" & n & "=" & i & vbCrLf) & ",X" & n & "=" & i '递归
Next
GETRESULT = Join(temp, vbCrLf)
Erase temp
End If
End If
End Function
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以上这段代码是求方程的解(网上转)
我觉得很好,但没怎么看懂
但是因为没有限定每个变量的取值范围,循环很耗时
我想假如能在里面加入每个变量允许的取值范围的话,循环时间能大大减少
但是我不知道,从那入手
大家帮我看看,就以x1取值只能为3到5
x2只能从2到4(我想其他的都一样)举例把
另外我想定义一个存放所有可能解的二维数组,比如 JIE(i,j)
i代表所有解的个数,j代表所有变量的个数
我想把所有的解存入这个数组中进行下一步的分析操作
以及定义一个变量来计算循环所用时间
请各位高手帮我出个主意吧
谢谢了