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标题:各位老师好!求助编辑一个大整数的快速乘除法可调用程序
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xianfajushi
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2022-02-16 21:20
ysr2857
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回复 341楼 xianfajushi
很好!您用的是啥语言编程的?看不懂啊!
2022-02-17 08:51
xianfajushi
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VB6数据类型Currency
数据类型Long
C++数据类型_int64
都是一样位数,可控制在18位进行运算
VB6没用到,用写出来你能用?
其实我的程序直接改为VB是很容易的,直接替换到数据类型即可,循环语句替换即可,程序内也没用到判断什么的,纯粹就是计算而已。
2022-02-17 11:42
ysr2857
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回复 343楼 xianfajushi
VB6数据类型Currency?这个不懂,有这个类型吗?
我不会,而和VB6是大不一样的。
2022-02-17 13:48
ysr2857
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回复 344楼 ysr2857
Private Sub Command1_Click()
Dim m, n
m = Trim(Text1): n = Trim(Text2)
ts = Timer
c = MbC4(Trim(m), Trim(n))
Text3 = c & "用时" & Timer - ts & "秒,有" & Len(c) & "位"
End Sub

Private Sub Command2_Click()
Text1 = ""
Text2 = ""
Text3 = ""

End Sub

Public Function MbC4(D1 As String, D2 As String) As String '快速乘法
        Dim l As Long, le As Long, le1 As Long, n As Long, r As Long, p As Long, q As Long, m As Byte
  Dim wr As Currency, w1 As Currency, wlr As Currency, wl1 As Currency, tr As Currency, t1 As Currency
  Dim pi As Currency, t As Currency, tr1 As Currency
        
  
Dim xr() As Currency, a As String
  a = Trim(D1)
  B = Trim(D2)
  
  X = Len(a) \ 6: Y = Len(B) \ 6
  a = String(Val(X * 6 + 6 - Len(a)), "0") & a
  B = String(Val(Y * 6 + 6 - Len(B)), "0") & B
  X = X + 1: Y = Y + 1
  sb1 = X + Y
  sb2 = Log(sb1) / Log(2)
  If InStr(sb2, ".") = 0 Then
  sb2 = sb2
  Else
  sb2 = Int(sb2) + 1
  End If
  sb = 2 ^ sb2
  a = String(Val(sb) * 6 - Len(a), "0") & a
  B = String(Val(sb) * 6 - Len(B), "0") & B
  
   ReDim x_(1 To sb): ReDim y_(1 To sb)
    For i1 = 1 To sb
    x_(i1) = Mid(a, (sb - i1 + 1) * 6 - 5, 6): y_(i1) = Mid(B, (sb - i1 + 1) * 6 - 5, 6)
    If Len(x_(i1)) < 6 Then
    x_(i1) = String(6 - Len(x_(i1)), "0") & x_(i1)
    ElseIf Len(y_(i1)) < 6 Then
    y_(i1) = String(6 - Len(y_(i1)), "0") & y_(i1)
    Else
    x_(i1) = x_(i1): y_(i1) = y_(i1)
    End If
   
      Next
      ReDim xr(0 To (Len(a) - 6) \ 6): ReDim yr(0 To (Len(B) - 6) \ 6): ReDim zr(0 To (Len(B) - 6) \ 6)
      
       If Len(a) = 6 Then
  xr(0) = a: yr(0) = B
  Else
    Dim I As Long, J As Long, mn As Long, lh As Long, k As Long
    '位序倒置
n = sb '求数组大小,其值必须是2的幂
lh = n / 2
    J = n / 2
    For I = 1 To n - 2


    Debug.Print I, J
    k = lh '下面是向右进位算法
Do
    If k > J Then Exit Do '高位是1吗
J = J - k '是的,高位置0
    k = k / 2 '准备次高位的权
Loop Until k = 0 '次高位的权若非0,则检查新的次高位
J = J + k '非则若最高位是0,则置1
   xr(I + 1) = x_(J + 1): yr(I + 1) = y_(J + 1)
    Next
    xr(0) = x_(1): xr(1) = x_(1 + sb / 2)
    yr(0) = y_(1): yr(1) = y_(1 + sb / 2)
  
     End If
  
  Dim xi() As Currency: Dim yi() As Currency: Dim zi() As Currency
  n = sb '求数组大小,其值必须是2的幂
m = 0
  l = 2
  pi = "3.141592653589793238"
  Do
  l = l + l
  m = m + 1
  Loop Until l > n
  n = l / 2
  ReDim xi(n - 1): ReDim yi(n - 1): ReDim zi(n - 1)

  l = 1
  Do
    le = 2 ^ l
    le1 = le / 2
    wr = 1
    wi = 0
    If l = 1 Then
    t = 0
    Else
    t = pi / le1
    End If
    w1r = Cos(t)
    w1i = -Sin(t)
    r = 0
  Do
    p = r
    Do
     q = p + le1
     
     tr = xr(q) * wr - xi(q) * wi
     ti = xr(q) * wi + xi(q) * wr
     tr1 = yr(q) * wr - yi(q) * wi
     ti1 = yr(q) * wi + yi(q) * wr
     
     
     xr(q) = xr(p) - tr
     xi(q) = xi(p) - ti
     xr(p) = xr(p) + tr
     xi(p) = xi(p) + ti
     
       yr(q) = yr(p) - tr1
      yi(q) = yi(p) - ti1
      yr(p) = yr(p) + tr1
      yi(p) = yi(p) + ti1
     xr(p) = Format(Val(xr(p)), "0.000000"): xi(p) = Format(Val(xi(p)), "0.000000")
     yr(p) = Format(Val(yr(p)), "0.000000"): yi(p) = Format(Val(yi(p)), "0.000000")
     
      p = p + le
  Loop Until p > n - 1


  wr2 = wr * w1r - wi * w1i
  wi2 = wr * w1i + wi * w1r
  wr = wr2
  wi = wi2
  r = r + 1
  Loop Until r > le1 - 1
  l = l + 1
  Loop Until l > m

  For I = 0 To n - 1 '仅输出模
   zr(I) = xr(I) * yr(I) - xi(I) * yi(I): zi(I) = xr(I) * yi(I) + xi(I) * yr(I)
    zr(I) = Format(Val(zr(I)), "0.000000"): zi(I) = Format(Val(zi(I)), "0.000000")
  

      's = s & "/" & zr(I)
      's1 = s1 & "/" & zi(I)
      Next
      
       J = sb
     
       ReDim x_(1 To sb): ReDim y_(1 To sb)
     For k = 1 To J
         n1 = n1 + 1
          ReDim Preserve x_(1 To n1)
        
         x_(n1) = zr(n1 - 1): y_(n1) = zi(n1 - 1)
         x_(n1) = Format(Val(x_(n1)), "0.000000"): y_(n1) = Format(Val(y_(n1)), "0.000000")
         
       Next
   
    '位序倒置
n = sb '求数组大小,其值必须是2的幂
lh = n / 2
    J = n / 2
   
    For I = 1 To n - 2


    Debug.Print I, J
    k = lh '下面是向右进位算法
Do
    If k > J Then Exit Do '高位是1吗
J = J - k '是的,高位置0
    k = k / 2 '准备次高位的权
Loop Until k = 0 '次高位的权若非0,则检查新的次高位
J = J + k '非则若最高位是0,则置1

xr(I + 1) = x_(J + 1): yr(I + 1) = y_(J + 1)
    'js = js & "/" & x_(J + 1)
    'js1 = js1 & "/" & y_(J + 1)
    Next
    'sx1 = "/" & x_(1) & "/" & x_(1 + sb / 2) & js
    'sy1 = "/" & y_(1) & "/" & y_(1 + sb / 2) & js1
   xr(0) = x_(1): xr(1) = x_(1 + sb / 2)
   yr(0) = y_(1): yr(1) = y_(1 + sb / 2)
   
   
   ns = Len(a) \ 6: Jn = ns
  
      
  

  ReDim zr(0 To ns - 1)

  m = 0
  l = 2
  pi = "3.141592653589793238"
  Do
  l = l + l
  m = m + 1
  Loop Until l > ns
  ns = l / 2
  ReDim xi(ns - 1): ReDim yi(ns - 1): ReDim zi(ns - 1)

  l = 1
  Do
    le = 2 ^ l
    le1 = le / 2
    wr = 1
    wi = 0
    If l = 1 Then
    t = 0
    Else
    t = -1 * pi / le1
    End If
    w1r = Cos(t)
    w1i = -Sin(t)
    r = 0
  Do
    p = r
    Do
     q = p + le1
     
     tr = xr(q) * wr - xi(q) * wi
     ti = xr(q) * wi + xi(q) * wr
     tr1 = yr(q) * wr - yi(q) * wi
     ti1 = yr(q) * wi + yi(q) * wr
     
     
     xr(q) = xr(p) - tr
     xi(q) = xi(p) - ti
     xr(p) = xr(p) + tr
     xi(p) = xi(p) + ti
     
       yr(q) = yr(p) - tr1
      yi(q) = yi(p) - ti1
      yr(p) = yr(p) + tr1
      yi(p) = yi(p) + ti1
     xr(p) = Format(Val(xr(p)), "0.000000"): xi(p) = Format(Val(xi(p)), "0.000000")
     yr(p) = Format(Val(yr(p)), "0.000000"): yi(p) = Format(Val(yi(p)), "0.000000")
      p = p + le
  Loop Until p > ns - 1


  wr2 = wr * w1r - wi * w1i
  wi2 = wr * w1i + wi * w1r
  wr = wr2
  wi = wi2
  r = r + 1
  Loop Until r > le1 - 1
  l = l + 1
  Loop Until l > m

  For I = 0 To ns - 1 '仅输出模
zr(I) = (xr(I) - yi(I)) / n
      zr(I) = Format(Val(zr(I) + 0.5), "0.000000")
     If InStr(zr(I), ".") = 0 Then
     s121 = zr(I)
     Else
     s121 = Left(zr(I), InStr(zr(I), ".") - 1)
      End If
      's0 = "/" & s121 & s0
      zr(I) = s121
      Next
      For i1 = 1 To Val(Jn - sb1 + 1)
      zr(sb1 + i1 - 2) = 0
      Next
      
     
     
      For i1 = 0 To n - 1
      If zr(i1) < 0 Then
      zr(i1) = "000000"
      ElseIf Len(zr(i1)) < 6 Then
      zr(i1) = String(6 - Len(zr(i1)), "0") & zr(i1)
      Else
      zr(i1) = zr(i1)
      End If
      
      's5 = s5 & "/" & zr(i1)
      
      If i1 = 0 Then
      
      s6 = Val(Left(zr(i1), Len(zr(i1)) - 6))
      If Len(s6) < 6 Then
      s6 = String(6 - Len(s6), "0") & s6
      Else
      s6 = s6
      End If
      s8 = Right(zr(i1), 6)
      ElseIf Val(zr(i1)) >= 0 Then
      s7 = Val(zr(i1)) + Val(s6)
    If Len(s7) = 6 Or Len(s7) = 12 Or Len(s7) = 18 Then
          s7 = s7
          Else
          s7 = String(24 - Len(s7), "0") & s7
         End If
      s10 = Right(s7, 6)
      s11 = s10 & s11
      If Len(s7) < 6 Then
      s7 = String(6 - Len(s7), "0") & s7
      ElseIf Len(s7) = 6 Then
      s6 = "000000"
      Else
      s7 = s7
      s6 = Val(Left(s7, Len(s7) - 6))
      End If
      Else
      s6 = s6
      End If
     
      Next
      s9 = s6 & s11 & s8
     
   
   s9 = qdqd0(Trim(s9))
   
     's2 = nifft(dxcx1(Trim(s)), dxcx1(Trim(s1)), Trim(sb1))
     
     's3 = nifft(Trim(sx1), Trim(sy1), Trim(sb1))
      MbC4 = s9
  End Function

  Private Function qdqd0(sa As String) As String
  a = sa
  Do While Left(a, 1) = "0"
  a = Mid(a, 2)
  Loop
  If a = "" Then
  a = 0
  Else
  a = a
  End If
  qdqd0 = a
  End Function

2022-02-17 15:30
ysr2857
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回复 345楼 ysr2857
   zr(I) = Val(xr(I) * yr(I) - xi(I) * yi(I)): zi(I) = Val(xr(I) * yi(I) + xi(I) * yr(I))

这一步溢出了
2022-02-17 15:42
xianfajushi
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初稿未优化
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a(8) = a(9)
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aa(8) = 777777777777777777
Console.WriteLine(a(7) & a(8) & a(9))
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[此贴子已经被作者于2022-2-17 16:24编辑过]

2022-02-17 15:56
ysr2857
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回复 347楼 xianfajushi
谢谢您!这是吧?
速度提高多少?
我不会 .
2022-02-18 02:09
xianfajushi
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我这种程序不废话,处理器处理数据有多快就有多快,也就是处理器的速度,不受多余的操作影响数据处理速度,每单元数据位是字符串型的18倍,字符串型每处理一位数都要进行多次操作,而我这程序操作处理就按每语句一次算处理后就是18位数据,而且我这程序绝对不会发生数据溢出错误问题。

[此贴子已经被作者于2022-2-18 12:54编辑过]

2022-02-18 08:30
xianfajushi
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不会其他语言没多大关系,关键是充分理解这种分治运算原理,明白了原理可以用任何语言写出来。
2022-02-18 10:52
快速回复:各位老师好!求助编辑一个大整数的快速乘除法可调用程序
数据加载中...
 
   



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