[第17期] 乘方[已总结]

winland

楼主的方法算出来同我的有些差别，很大的数会差2，用下面的例子算会差1。用我的方法是6146，楼主的方法是6145。我是用下面的方法把所有的全列出来验证的。

Function powercount3(dinput As Double) As Double
    Dim i As Double
    Dim tmp As String
    Dim j As Long
   
    j = 1
    For i = dinput To 2 Step -1
        tmp = maxpower3(i)
        If tmp <> "0=" Then
            i = Mid(tmp, 1, InStr(tmp, "=") - 1)
            Cells(j, 26) = i
            Cells(j, 27) = Mid(tmp, InStr(tmp, "="))
            j = j + 1
        Else
            Exit For
        End If
    Next i
    powercount3 = "OK"
End Function

Function maxpower3(dinput As Double) As String
    Dim tmp As Double
    Dim tmpStr As String
    Dim tmpL As Double
    tmpStr = ""
   
    For i = 2 To dinput
        tmpL = Int(dinput ^ (1 / i))
        If tmpL = 1 Then
            Exit For
        Else
            'when the difference is very small, some strange thing happen
            '------------------------------------------------------------
            If tmpL ^ i > dinput Then
                tmpL = tmpL - 1
            End If
            If (tmpL + 1) ^ i = dinput Then
                tmpL = tmpL + 1
            End If
            '-----------------------------------------------------------
            If tmpL ^ i > tmp Then
                tmp = tmpL ^ i
               
                tmpStr = tmpL & "^" & i
            End If
        End If
    Next i
    maxpower3 = tmp & "=" & tmpStr
End Function
Sub mmm()
    MsgBox powercount3(33666665)
End Sub

northwolves

总结：

本期题目其实难度不大，关键是要特别细心，一不小心就会无中生有或丢三落四。

大家的解法和思路基本都正确。但参赛者比较少，估计有些朋友被题目的大数字吓倒了。

fjlhgs

出错的原因是应为vba开方运算" ^ "本身存在误差，vba在计算int(125^(1/3))时计算出的值为4,实际为5。据此我自定义了2个函数来消除误差，附件为修正后的程序，程序速度稍慢。33666665用我修正后的程序计算结果为6141。

 

chrisfang

powercount(33666665) 用我的算出来也是6146。

还没研究版主的以及版主推荐的算法，过会儿看看结果的误差发生在哪里。

northwolves

QUOTE:

以下是引用winland在2006-11-20 23:40:40的发言：

楼主的方法算出来同我的有些差别，很大的数会差2，用下面的例子算会差1。用我的方法是6146，楼主的方法是6145。我是用下面的方法把所有的全列出来验证的。

Function powercount3(dinput As Double) As Double
    Dim i As Double
    Dim tmp As String
    Dim j As Long
   
    j = 1
    For i = dinput To 2 Step -1
        tmp = maxpower3(i)
        If tmp <> "0=" Then
            i = Mid(tmp, 1, InStr(tmp, "=") - 1)
            Cells(j, 26) = i
            Cells(j, 27) = Mid(tmp, InStr(tmp, "="))
            j = j + 1
        Else
            Exit For
        End If
    Next i
    powercount3 = "OK"
End Function

Function maxpower3(dinput As Double) As String
    Dim tmp As Double
    Dim tmpStr As String
    Dim tmpL As Double
    tmpStr = ""
   
    For i = 2 To dinput
        tmpL = Int(dinput ^ (1 / i))
        If tmpL = 1 Then
            Exit For
        Else
            'when the difference is very small, some strange thing happen
            '------------------------------------------------------------
            If tmpL ^ i > dinput Then
                tmpL = tmpL - 1
            End If
            If (tmpL + 1) ^ i = dinput Then
                tmpL = tmpL + 1
            End If
            '-----------------------------------------------------------
            If tmpL ^ i > tmp Then
                tmp = tmpL ^ i
               
                tmpStr = tmpL & "^" & i
            End If
        End If
    Next i
    maxpower3 = tmp & "=" & tmpStr
End Function
Sub mmm()
    MsgBox powercount3(33666665)
End Sub

你是对的，看来符点运算是存在很多问题。

northwolves

找见问题所在了：

Private Const p1 As String = "{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47}"
Private Const p2 As String = "{6,10,14,15,21,22,25,26,33,34,35,38,39,46}"

Function powercount(ByVal r As Range) As Long
powercount = Evaluate("SUMPRODUCT(INT(" & r & "^(1/" & p1 & ")))-SUMPRODUCT(INT(" & r & "^(1/" & p2 & ")))")  '前者15个质数，1被计数15次，后者14个质数，1被计数14次，正好抵消
End Function

Private Const p2 As String = "{6,10,14,15,21,22,26,33,34,35,38,39,46}" '1，2的误差发生在2^25,3^25 处，由于指数完全平方数时不会被重复计数，故25不应计数在内

Function powercount(ByVal r As Range) As Long
powercount = Evaluate("SUMPRODUCT(INT(" & r & "^(1/" & p1 & ")))-SUMPRODUCT(INT(" & r & "^(1/" & p2 & ")))") - 1 '前者15个质数，1被计数15次，后者13个质数，1被计数13次，总计多计1次，故减1
End Function

Function maxpower(ByVal r As Range) As String
Dim largest As Double, power As Byte, p3 As String
largest = Evaluate("MAX(INT(" & r & "^(1/" & p1 & "))^" & p1 & ")")
p3 = largest & "^(1/" & p1 & "))"
power = Evaluate("1/Max((INT(" & p3 & "=" & p3 & "*(1/" & p1 & "))")
maxpower = largest & "=" & largest ^ (1 / power) & "^" & power
End Function

fjlhgs

QUOTE:

以下是引用fjlhgs在2006-11-21 9:47:26的发言：

出错的原因是应为vba开方运算" ^ "本身存在误差，vba在计算int(125^(1/3))时计算出的值为4,实际为5。据此我自定义了2个函数来消除误差，附件为修正后的程序，程序速度稍慢。33666665用我修正后的程序计算结果为6141。

 

[此贴子已经被作者于2006-11-23 16:58:35编辑过]