[Excel 程序开发] [开_102]一道趣味算术题

一道有趣的算术题（见于北京市某年小学奥数竞赛）

先看几个数：312132,231213,41312432,23421314

可以看到以上字串有以下特点： 2个1 之间有1个数，2个2 之间有2个数，2个3之间有3个数...
试找出由(1,1,2,2,3,3,4,4,5,5,6,6,..N,N) 2*N(N<=8) 个数组成的字符串满足上述条件的所有组合,即2个1 之间有1个数，2个2 之间有2个数，2个3之间有3个数...2个N之间有N个数，通过点击上面的“全部列出”按钮将其列到前2*N列中

可以证明 N 只能是 4K,4K+3 的形式.

N=1,2,5,6,9时无解
N=3:   312132,231213
N=4:   41312432,23421314
N=7;   52个解
N=8:   300个解

截至日期 2006-8-1，只要代码不雷同，结果正确，计算快速，算法巧妙均有加分机会。

有点麻烦,得仔细琢磨下先.

俺对数学题也晕.
穷举,心里踏实一些.不再去想优化了,格式上和版主的要求有点差异.权做抛砖了.

VSEFUarA.rar (16.75 KB, 下载次数: 118)

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[VBA16]一道趣味算术题

数学题 , interesting

沒有基本原理 [em04]

这样行吗

djqNCCjs.rar (72.61 KB, 下载次数: 159)

无解.

假设1,2,3,4,...n 其中较小的一个位于 a1,a2,a3,a.....,an,则较大的一个分别位于a1+2,a2=3,a3+4,...an+n+1

而且:

a1,a2,a3,a.....,an,   a1+2,a2=3,a3+4,...an+n+1 为1,1,2,2,3,3,...n,n 的一个全排列

则(a1+a2+...+an  )+ (a1+2 +a2+3+a3+4+...an+n+1)=2*(1+2+3+...+n)=n*(n+1)

所以2(a1+a2+...+an)+ (2+3+...+n+1)=n*(n+1)

a1+a2+...+an=n(n+1)/4

所以,n一定是4k 或 4k+3 的形式.

另:

111222333444555666777888999 倒可以满足,呵呵:

3 4 7 9 3 6 4 8 3 5 7 4 6 9 2 5 8 2 7 6 2 5 1 9 1 8 1

Almost-Langford numbers , written by pgc01.

Option Explicit
Option Base 1
Const lCol As Long = 1 ' Writes in Column A

Sub AlmostLangford()
Dim iDigits(), iNDigits As Integer, lRow As Long
   
    For iNDigits = 2 To 18 Step 2
        ReDim iDigits(iNDigits)
        Call AlmostLangfordN(iDigits, lRow)
    Next

End Sub

Sub AlmostLangfordN(ByRef iDigits(), ByRef lRow As Long)
Dim iPosFree As Integer, iDigitsTmp(), iDigit As Integer
Dim iNDigits As Integer, i As Integer

    iNDigits = UBound(iDigits)
    Do While True
        'Checks if last free position was filled
        If iPosFree <> 0 Then If IsEmpty(iDigits(iPosFree)) Then Exit Sub
        'Determines next free position
        For iPosFree = iPosFree + 1 To iNDigits
            If IsEmpty(iDigits(iPosFree)) Then Exit For
        Next
        If iPosFree = iNDigits + 1 Then Exit Sub

        ' Tries all digits
        For iDigit = 0 To 9
            If iPosFree + iDigit + 1 > iNDigits Then Exit For
            ' Checks if digit already used
            For i = 1 To iNDigits
                If iDigits(i) = iDigit And Not IsEmpty(iDigits(i)) Then Exit For
            Next
            ' Cannot start with zero
            If i > iNDigits And Not (iPosFree = 1 And iDigit = 0) Then
                If IsEmpty(iDigits(iPosFree + iDigit + 1)) Then
                    ' Initialises idigitstmp
                    ReDim iDigitsTmp(iNDigits)
                    For i = 1 To iNDigits
                        iDigitsTmp(i) = iDigits(i)
                    Next
                    iDigitsTmp(iPosFree) = iDigit
                    iDigitsTmp(iPosFree + iDigit + 1) = iDigit
                   
                    ' Checks if array is full (Almost-Langford number)
                    For i = 1 To iNDigits
                        If IsEmpty(iDigitsTmp(i)) Then Exit For
                    Next
                    If i > iNDigits Then
                        lRow = lRow + 1
                        Cells(lRow, lCol) = "'" & Join(iDigitsTmp, "")
                    Else
                        Call AlmostLangfordN(iDigitsTmp, lRow) 'iDigit + 1
                    End If
                End If
            End If
        Next iDigit
    Loop

End Sub

兄台的功力实在令人敬佩.有兴趣把3个N的也做一做. n mod 9={0,1,8} 时有解.