回复 13楼 yuma
大整数的加减法容易办到，直接模仿手工计算就行，而且可以从高位算起，当然从低位算起再道序也行，速度很快，就是直接模仿手工竖式算法就行，注意进位（或借位）的问题。乘法和除法我也仅会模仿手工计算的方法，效率低速度慢，不讲了。
把大整数的加法和减法的可调用程序发一下，仅供参考：
加法程序：
Public Function MPC1(D1 As String, D2 As String) As String 'jiafa
Dim x, Y '两数长度
If Len(D1) >= Len(D2) Then
D4 = String(Len(D1) - Len(D2), "0") & D2
d3 = D1
Else
D4 = D2
d3 = String(Len(D2) - Len(D1), "0") & D1
End If
x = Len(d3): Y = Len(D4)
Dim a() As Integer, B1() As Integer, C1() As Integer, E1() As Integer
ReDim a(1 To x)
ReDim B1(1 To Y)
ReDim C1(1 To x)
ReDim E1(1 To x)
Dim I, J, C2, CJ, JW
For J = Y To 1 Step -1 'D2
JW = 0 '进位清0
B1(J) = Mid$(D4, J, 1) '每位数
For I = x To 1 Step -1 'D1
a(I) = Mid$(d3, I, 1) '每位数
C1(I) = a(I) + B1(I) + JW '计算jia
JW = C1(I) \ 10
E1(I) = C1(I) Mod 10
Next
Next
For r = 1 To x
If JW = 0 Then
MPC1 = MPC1 & E1(r)
Else
jc = jc & E1(r)
MPC1 = JW & jc
End If
Next
End Function
减法程序（必须大的减去小的不输出负值）：
Public Function MPC(D1 As String, D2 As String) As String ';jianfaqi
Dim x, Y ';两数长度
If Len(D1) >= Len(D2) Then
D4 = String(Len(D1) - Len(D2), "0") & D2
d3 = D1
Else
D4 = D2
d3 = String(Len(D2) - Len(D1), "0") & D1
End If
x = Len(d3): Y = Len(D4)
Dim a() As Integer, B1() As Integer, C1() As Integer, E1() As Integer
ReDim a(1 To x)
ReDim B1(1 To Y)
ReDim C1(1 To x)
ReDim E1(1 To x)
Dim I, J, C2, CJ, JW
For J = Y To 1 Step -1 ';D2
JW = 1 ';yu jie weichuzhi
B1(J) = Mid(D4, J, 1) ';每位数
For I = x To 1 Step -1 ';D1
a(I) = Mid(d3, I, 1) ';每位数
C1(I) = 10 + a(I) - B1(I) - 1 + JW ';计算jia
JW = C1(I) \ 10
E1(I) = C1(I) Mod 10
Next
Next
For r = 1 To x
MPC = MPC & E1(r)
For I = 1 To Len(MPC)
If Not Mid(MPC, I, 1) = "0" Then
Exit For
End If
Next
strtmp = Mid(MPC, I)
If Len(strtmp) = 0 Then
MPC = "0"
Else
MPC = strtmp
End If
Next
End Function
