输出余数的除法结果：123456789876543/987654321=124999/987405864，
输出带小数点的除法：123456789876543/987654321=124999.9997484372。

我的除法只算到整数部分，要得到小数点后的商值被除数末尾必须补0，精确几位补几个0.

当余数为0的时候不要输出/0，因为判断是否整除的时候是以有没有/号为标准的，这一点要注意！

下面是网上找到的用vc编程的fft快速高精度乘法程序，我看不懂，请看看，翻译成vb可调用程序，谢谢！
#include<bits/stdc++.h>
using namespace std;
//complex是stl自带的定义复数的容器
typedef complex<double> cp;
#define N 2097153
//pie表示圆周率π
const double pie=acos(-1);
int n;
cp a[N],b[N];
int rev[N],ans[N];
char s1[N],s2[N];
//读入优化
int read(){
    int sum=0,f=1;
    char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return sum*f;
}
//初始化每个位置最终到达的位置
{
    int len=1<<k;
    for(int i=0;i<len;i++)
    rev[i]=(rev[i>>1]>>1)|((i&1)<<(k-1));
}
//a表示要操作的系数，n表示序列长度
//若flag为1，则表示FFT，为-1则为IFFT(需要求倒数）
void fft(cp *a,int n,int flag){
    for(int i=0;i<n;i++)
    {
     //i小于rev[i]时才交换，防止同一个元素交换两次，回到它原来的位置。
      if(i<rev[i])swap(a[i],a[rev[i]]);
    }
    for(int h=1;h<n;h*=2)//h是准备合并序列的长度的二分之一
    {
    cp wn=exp(cp(0,flag*pie/h));//求单位根w_n^1
     for(int j=0;j<n;j+=h*2)//j表示合并到了哪一位
     {
      cp w(1,0);
       for(int k=j;k<j+h;k++)//只扫左半部分，得到右半部分的答案
       {
         cp x=a[k];
         cp y=w*a[k+h];
         a[k]=x+y;  //这两步是蝴蝶变换
         a[k+h]=x-y;
         w*=wn; //求w_n^k
       }
     }
     }
     //判断是否是FFT还是IFFT
     if(flag==-1)
     for(int i=0;i<n;i++)
     a[i]/=n;
}
int main(){
    n=read();
    scanf("%s%s",s1,s2);
    //读入的数的每一位看成多项式的一项，保存在复数的实部
    for(int i=0;i<n;i++)a[i]=(double)(s1[n-i-1]-'0');
    for(int i=0;i<n;i++)b[i]=(double)(s2[n-i-1]-'0');
    //k表示转化成二进制的位数
    int k=1,s=2;
     while((1<<k)<2*n-1)k++,s<<=1;
    init(k);
    //FFT 把a的系数表示转化为点值表示
    fft(a,s,1);
    //FFT 把b的系数表示转化为点值表示
    fft(b,s,1);
    //FFT 两个多项式的点值表示相乘
    for(int i=0;i<s;i++)
    a[i]*=b[i];
    //IFFT 把这个点值表示转化为系数表示
    fft(a,s,-1);
    //保存答案的每一位(注意进位）
    for(int i=0;i<s;i++)
    {
    //取实数四舍五入，此时虚数部分应当为0或由于浮点误差接近0
    ans[i]+=(int)(a[i].real()+0.5);
    ans[i+1]+=ans[i]/10;
    ans[i]%=10;
    }
    while(!ans[s]&&s>-1)s--;
    if(s==-1)printf("0");
    else
    for(int i=s;i>=0;i--)
    printf("%d",ans[i]);
    return 0;
}

给出一个超大整数快速求商的原理图，您是否能编程vb版程序：
对于超大整数步骤是这样：
初始：c=0，i>=0
1，满足被除数>=2^i*除数，得到最大的i0.c=c+2^i0.
新的被除数=被除数-2^i0*除数。
2，满足新被除数>=2^i*除数的最大的i1.c=c
3，若2^0=1>=2^i的i不存在，则商c=c+2^0.
我是这么理解的。
速度取决于求2^i的速度，以及比较大小的速度，行不行？请指导！
（我的理解第一步和第二步都是迭代，第三步i已经等于0，无需迭代仅是个判断。）

[此贴子已经被作者于2020-2-19 06:24编辑过]

我觉得i不用从0开始算就可以得到i0及i1，方法：
被除数的位数减掉除数的位数，设差为x，x乘以0.3010就是i，因为log2=0.3010，而x*log2是2^i的位数，此时结果要小一点，比如减1或2，再迭代就可以减少迭代次数，（试了一下，不对，太大，应该令x1=x+1，迭代，当x1*log2=x时的x1的值。）
下一步类似，求出i1的近似值，要小于i1，就是新被除数的位数减掉除数的位数 ，再用上面的方法求出i1的近似值再迭代，直到i等于0为止。
当然，若被除数的位数和除数的位数的比值是1，那就可以从0开始算了。
我先试一下，如果成立把程序发一下，请您优化一下。

[此贴子已经被作者于2020-2-19 10:01编辑过]

这种方法不能实现，速度也许快，弄不准确，第一步就无法调整，多乘一次2则过大，远大于实际，少乘一次又过小。请把我发在84楼的vc程序翻译成vb程序吧，谢谢！欢迎指导！欢迎沟通！

大致知道你确定大素数的思想了：n是素数，和n的互质数有n-1个，找到一个和n-1互质的数c，求出逆元d，用n和c对明文a加密得到密文b，用n和d对b解密得到a1，如果a1=a则说明n是素数，反之不是。
问题是这个理论能否成立呢？我感觉以往做rsa时，有些合数也可以正确加解密的。我是用Miller-Rabin检测大数素性的，但是要找到P、Q两个随机大素数很费时（我通常使用的P、Q都在十进制120位以上），用时多在10分钟以上，为了加快速度，我不断减少Miller-Rabin算法的检测样本数，我发现在很多时候都能实现加解密，Miller-Rabin算法是无法确保得到大质数的，我总感觉其中好多次在P、Q不是质数时也通过了加解密。
抽空用c做下。
我做的大数除法没有你快，我的除法过份依赖乘法和加法，但用我的加减乘配合你的除法可以大幅度提高运算速度，代码在下面工程里，你可以参考下：

我提供的是工程目录及相关文件的压缩包，解压后就可用vb打开这个工程了，你不会连解压缩文件都没有吧！全部代码如下：

Private Function zhengchuqyushu(sa As String) As String
'获取余数
If InStr(sa, "/") = 0 Then

 zhengchuqyushu = 0

 Else

 zhengchuqyushu = Mid(sa, InStr(sa, "/") + 1)

 End If
End Function


Private Function zhengchuqy(sa As String) As String
'获取商
If InStr(sa, "/") = 0 Then

 zhengchuqy = sa

 Else

 zhengchuqy = Left(sa, InStr(sa, "/") - 1)

 End If
End Function


Public Function MBBC(d1 As String) As String 'kai pingfang
If Len(d1) < 10 Then

 jss = Int(Sqr(d1))

 JW = Val(d1) - (jss) ^ 2
  If JW = 0 Then
  MBBC = jss
  Else
  MBBC = jss & "/" & JW
    End If

 Else
Dim x 'shuju changdu
x = Len(d1) \ 4
d2 = String(4 - Len(d1) + 4 * x, "0") & d1
Dim a() As String
ReDim a(4 To 4 * x + 4)
Dim b() As String
ReDim b(2 To 2 * x)
Dim i, j, js
  For i = 4 To 4 * x + 4 Step 4
  
a(i) = Mid(d2, i - 3, 4)
js = Int(Sqr(Val(a(4) & a(8))))

 JW = Val(a(4) & a(8)) - (js) ^ 2

 Next
   j = 4
   Do While j <= 2 * x
   
   jws = MPC1(JW & "0000", a(2 * j + 4))
   If MBJC(Trim(jws), MbC(Trim(js), 200)) <= 0 Then
    b(j) = "00"
    Else
    jwc = Left(jws, Len(jws) - Len(MbC(Trim(js), 200)) + 2) \ Left(MbC(Trim(js), 200), 2)
    If Len(jwc) > 2 Then
     b(j) = 99
     Else
     b(j) = jwc
     End If
   
     
     Do While MBJC(Trim(jws), MbC(MPC1(b(j), MbC(Trim(js), 200)), b(j))) = -1
     
     b(j) = b(j) - 1
     
               Loop
          End If
          JW = MPC(Trim(jws), MbC(MPC1(MbC(200, Trim(js)), b(j)), b(j)))
       
     js = MPC1(MbC(Trim(js), 100), Trim(b(j)))
     
       
   j = j + 2
   If JW = 0 Then
       
   MBBC = js
   Else
   MBBC = js & "/" & JW
   End If
   Loop
   

 End If
End Function

Public Function MBJC(d1 As String, d2 As String) As String ';bijiao
If Len(d1) <= 10 And Len(d2) <= 10 Then
If Val(d1) > Val(d2) Then
MBJC = 1
Else
If Val(d1) = Val(d2) Then
MBJC = 0
Else
MBJC = -1
End If
End If
Else

If Len(d1) > Len(d2) Then
MBJC = 1
Else
If Len(d1) < Len(d2) Then
MBJC = -1
Else
If Len(d1) = Len(d2) Then

 Dim x, Y

 x = Len(d1) \ 4: Y = Len(d2) \ 4

 Dim a() As String, b() As String

 ReDim a(4 To 4 * x + 4)

 ReDim b(4 To 4 * Y + 4)

 If Val(Left(d1, Len(d1) - 4 * x)) > Val(Left(d2, Len(d2) - 4 * Y)) Then
  MBJC = 1
  Else
  If Val(Left(d1, Len(d1) - 4 * x)) < Val(Left(d2, Len(d2) - 4 * Y)) Then
  MBJC = -1
  Else
  For i = 4 To 4 * x Step 4
  a(i) = Mid(d1, Len(d1) - i + 1, 4)
  b(i) = Mid(d2, Len(d2) - i + 1, 4)
  Next
  j = 4 * x
  Do While a(j) = b(j) And j >= 8
  
  j = j - 4
     Loop
     
     
   If Val(a(j)) - Val(b(j)) > 0 Then
   MBJC = 1
   Else
   If Val(a(j)) - Val(b(j)) < 0 Then
   MBJC = -1
   Else
   MBJC = 0
   End If
   
  End If
  
  
  

 End If
End If
End If
End If
End If
End If
End Function

Public Function MCC(d1 As String, d2 As String) As String '程序
If Len(d1) < Len(d2) Then
   MCC = "0" & "/" & d1
   Else
   If Len(d1) < 9 Then
    MCC = Val(d1) \ Val(d2) & "/" & Val(d1) - (Val(d1) \ Val(d2)) * Val(d2)
     If Mid(MCC, InStr(MCC, "/") + 1) = 0 Then
  MCC = Left(MCC, InStr(MCC, "/") - 1)

 Else
MCC = MCC
End If
    
    Else
    
   Dim x ';fen duan changdu
   x = Len(d1)
    
     
    
     Dim a() As String
      ReDim a(1 To x)  '程序
      For i = 1 To x Step 1  '程序
       a(i) = Mid(d1, i, 1)
        
       
       Next i
      Dim b() As String
      JW = 0
     ReDim b(1 To x)
     For j = 1 To x Step 1
    b(j) = Val(JW & a(j)) \ Val(d2)
      JW = Val(JW & a(j)) - Val(b(j)) * Val(d2)
       Next j
       For r = 1 To x
       If JW = 0 Then
          MCC = MCC & b(r)
          Else
          CJ = CJ & b(r)
          MCC = CJ & "/" & JW
       
    End If
    
    For i = 1 To Len(MCC)
   If Not Mid(MCC, i, 1) = "0" Then
       Exit For
   End If
Next
strtmp = Mid(MCC, i)

 If Len(strtmp) = 0 Then

 MCC = "0"

 Else
MCC = strtmp
End If
    
   Next
   
   End If
     
     End If
   
End Function

 
Public Function mComp(ByVal d1 As String, ByVal d2 As String) As Integer
  '大数比较函数，-1:d1<d2  0:d1=d2 1:d1>d2
'  mComp = Val(Left(MPC(D1, D2), 2))
'  If mComp <> 0 Then mComp = Int(mComp / Abs(mComp))
'  Exit Function
  Dim i As Integer, j As Integer
  mFormat d1
  mFormat d2
  i = Len(d1)
  j = Len(d2)
  mComp = 1
  If i = j Then
    If d1 = d2 Then mComp = 0
    If d1 < d2 Then mComp = -1
  Else
    If i < j Then mComp = -1
  End If
End Function

Public Function mFormat(d As String) As Integer
  '格式化数据，将数据格式化为纯数字，非数字字符替换为0，大数为负数则去掉负号并返回-1，否则返回1
  Dim i As Integer, a As String, b As String
  a = Trim(d)
  d = ""
  mFormat = 1
  If Left(a, 1) = "-" Then mFormat = -1
  For i = 1 To Len(a)
    '本循环将大数中非数字字符用数字0代替，如-123a456=01230456
    b = Mid(a, i, 1)
    If b >= "0" And b <= "9" Then
      d = d & b
    Else
      d = d & "0"
    End If
  Next
  For i = 1 To Len(d)
    If Mid(d, i, 1) > 0 Then Exit For
  Next
  d = Right(d, Len(d) + 1 - i) '消前导0，如0000123456=123456
  If d = "" Then d = "0"
End Function

Public Function MCC1(d1 As String, d2 As String) As String '去问号
Dim ss As String, s As Integer
'MCC1 = MCC3(d1, d2)
'Exit Function             '去掉这两句的单引号就会用我的大数除法，速度慢，需要300秒，注释掉后就会用题主的除法算法，20多秒可以完成
ss = MBJC(d1, d2)
If Val(ss) = -1 Then
  MCC1 = "0" & "/" & d1
Else
  If Val(ss) = 0 Then
    MCC1 = 1
  Else
    If Len(d1) = Len(d2) Then
      s = Val(Left(d1, 1)) \ Val(Left(d2, 1))
      Do While MBJC(MbC(Trim(s), Trim(d2)), d1) = 1
        s = s - 1
      Loop
      If MBJC(MbC(Trim(s), Trim(d2)), d1) = 0 Then
        MCC1 = s
      Else
        MCC1 = s & "/" & MPC(Trim(d1), MbC(Trim(s), Trim(d2)))
      End If
    Else
      If Len(d2) < 9 Then
        MCC1 = MCC(d1, d2)
      Else
        Dim x, Y '变问号了咋
        x = Len(d1): Y = Len(d2)
        Dim JW, jcc, jss, jcs
        Dim a() As String, b() As String
        ReDim a(1 To x)
        ReDim b(1 To Y)
        For i = 1 To x
          a(i) = Mid(d1, i, 1)
        Next
        For j = 1 To Y
          b(j) = Mid(d2, j, 1)
        Next
        jcc = Val(a(1) & a(2)) \ Val(b(1) & b(2))
        jss = MbC(Trim(jcc), d2)
        For i1 = 1 To Y
          jws = jws & a(i1)
        Next
        Do While MBJC(Trim(jws), Trim(jss)) = -1
          jcc = jcc - 1
          jss = MbC(Trim(jcc), d2)
        Loop
        JW = MPC(Trim(jws), Trim(jss))
        z = x - Y
        Dim c() As String
        ReDim c(1 To z)
        For s = 1 To z
          If MBJC(JW & a(s + Y), d2) = -1 Then
            c(s) = "0"
          Else
            jwc = Val(Left(JW & a(s + Y), 3)) \ Val(Left(d2, 2))
            If Len(jwc) > 1 Then
              c(s) = "9"
            Else
              c(s) = jwc
            End If
            Do While MBJC(JW & a(s + Y), MbC(Val(c(s)), d2)) = -1
              c(s) = Right(10000 + Val(c(s) - 1), 1)
            Loop
          End If
          JW = MPC(JW & a(s + Y), MbC(Val(c(s)), d2))
          jcc = jcc & c(s)
        Next s
        If JW = 0 Then
          MCC1 = jcc
        Else
          MCC1 = jcc & "/" & JW
        End If
        For i = 1 To Len(MCC1)
          If Not Mid(MCC1, i, 1) = "0" Then Exit For
        Next
        strtmp = Mid(MCC1, i)
        If Len(strtmp) = 0 Then
          MCC1 = "0"
        Else
          MCC1 = strtmp
        End If
      End If
    End If
  End If
End If
End Function

Public Function MCC3(ByVal d1 As String, ByVal d2 As String) As String
  '格式转换，将我的大数除法结果转换为题主设定的“商/余数”格式
  Dim a As String, b As String
  a = MCC5(d1, d2)
  b = MPC(d1, MbC(d2, a))
  If Val(b) > 0 Then a = a & "/" & b
  MCC3 = a
End Function

Public Function MCC5(ByVal d1 As String, ByVal d2 As String) As String
  '大数除法，d1/d2，不处理负数，参数中非数字字符按0处理
  Dim i As Long, j As Long, k As Long, a As String, b As String, c As String
  Dim l1 As Integer, l2 As Integer, l As Integer
  i = mComp(d1, d2)
  If i < 1 Then
    MCC5 = i + 1
    Exit Function  '返回被除数小于或等于除数的商
  End If
  MCC5 = d1
  If Val(d2) < 2 Then Exit Function '如果除数为0或1则直接把被除数作为结果返回（除0不给出错误）
  a = ""
  If Len(d2) < 9 Then
    k = Val(d2)
    j = 0
    For i = 1 To Len(d1)
      If j > 100000000 Then
        a = a & Int(j / k)
        j = j Mod k
      Else
        If a <> "" And j < k Then a = a & "0"
      End If
      j = j * 10 + Val(Mid(d1, i, 1))
    Next
    a = a & Int(j / k)
  Else
    b = ""
    a = ""
    i = 1
    While i <= Len(d1)
      j = Len(b)
      b = b & Mid(d1, i, Len(d2) + 1 - j)  '多加一位是确保b>d2
      j = Len(b) - j
      i = i + j
      l = 0
      l1 = 0
      l2 = 100
      If mComp(b, d2) >= 0 Then
        While l2 > l1 + 1
          l = Int(l2 + l1) / 2
          c = MbC(d2, l)
          If mComp(b, c) < 0 Then
            l2 = l
          Else
            l1 = l
          End If
        Wend
        b = MPC(b, MbC(d2, l1))
        If Val(Left(b, 2)) = 0 Then b = "" '获取余数
      End If
      c = Trim(l1)
      If l1 = 0 Then c = ""
      If a <> "" Then
        For k = 1 To j - Len(c)
          a = a & "0"
        Next
      End If
      a = a & c
    Wend
  End If
  MCC5 = a
End Function
Public Function MbC(ByVal d1 As String, ByVal d2 As String) As String
  '大数乘法，d1*d2，不处理负数，参数中非数字字符按0处理
  Dim i As Integer, j As Long, k As Integer, a As String, b As String, Y As String
  Y = ""
  a = "0"
  For i = Len(d1) To 1 Step -1
    j = 0
    b = ""
    For k = Len(d2) To 1 Step -1
      j = Val(Mid(d1, i, 1)) * Val(Mid(d2, k, 1)) + j
      b = (j Mod 10) & b
      j = Int(j / 10)
    Next
    If j > 0 Then b = j & b
    a = MPC1(a, b & Y)
    Y = Y & "0"
  Next
  MbC = a
End Function

Public Function MPC(ByVal d1 As String, ByVal d2 As String) As String
  '大数减法d1-d2，如果d2>d1则交换，非法字符当字符0处理，不识别负数
  Dim a As String, b As String, c As String, i As Integer
  If mComp(d1, d2) < 0 Then                     '确保被减数大于减数
    MPC = MPC(d2, d1)                           '这里可根据需要输出负数
    Exit Function
  End If
  c = "9876543210"
  a = ""
  b = ""
  For i = 1 To Len(d2)
    b = b & Mid(c, Val(Mid(d2, i, 1)) + 1, 1)   '对减数按位取反
  Next
  For i = Len(d2) + 1 To Len(d1)
    b = "9" + b
  Next
  b = MPC1(b, "1")                              '调整该减数为对应十进制补数
  b = MPC1(d1, b)
  a = Right(b, Len(d1))
  For i = 1 To Len(a)
    If Mid(a, i, 1) <> "0" Then Exit For
  Next
  a = Right(a, Len(a) + 1 - i)                  '消前导0
  If a = "" Then a = "0"
  MPC = a
End Function

Public Function MPC1(ByVal d1 As String, ByVal d2 As String) As String
  '大数加法d1+d2,函数不识别参数的合法性,参数中有非法字符当作0处理,不识别负数
  Dim l1 As Integer, l2 As Integer, j As Integer, a As Integer, b As Integer
  l1 = Len(d1)
  l2 = Len(d2)
  j = 0
  While l1 + l2 + j > 0
    a = 0
    b = 0
    If l1 > 0 Then
      a = Val(Mid(d1, l1, 1))
      l1 = l1 - 1
    End If
    If l2 > 0 Then
      b = Val(Mid(d2, l2, 1))
      l2 = l2 - 1
    End If
    j = a + b + j
    MPC1 = (j Mod 10) & MPC1
    j = Int(j / 10)
  Wend
End Function


 Private Function zzxc(sa As String, sb As String) As String

 Dim a, b, c, d, r
  a = Trim(sa)
  b = Trim(sb)
  If Len(a) < 10 And Len(b) < 10 Then
  
  If Val(a) > Val(b) Then
     c = a
     d = b
  Else
     c = b
     d = a
  End If

 Do Until Val(c) Mod Val(d) = 0
     r = c Mod d
     c = d
     d = r
  Loop
  
  Else
  
  If MBJC(Trim(a), Trim(b)) >= 1 Then
  c = a
  d = b
  Else
  c = b
  d = a
  End If
  Do Until zhengchuqyushu(MCC1(Trim(c), Trim(d))) = 0
  r = zhengchuqyushu(MCC1(Trim(c), Trim(d)))
  c = d
  d = r
  Loop
  End If

 
  
  zzxc = d
  
End Function


 Private Function qniyuan(sa As String, sb As String) As String

 Dim n, p, a, b, c, d, r
  n = Trim(sa)
  p = Trim(sb)
  a = 1
  b = 0
  c = 0
  d = 1
  If Len(n) < 10 And Len(p) < 10 Then
  
  If Val(n) > Val(p) Then
     m = n
     q = p
     s1 = 1
  Else
     m = p
     q = n
     s1 = 0
  End If

 Do Until Val(m) Mod Val(q) = 0
    s = m \ q
     r = m Mod q
     s1 = s1 + 1
     If s1 Mod 2 = 1 Then
     a = a
     b = a * s + b
     c = c
     d = c * s + d
     Else
     b = b
     a = a + b * s
     d = d
     c = c + d * s
     End If
     m = q
     q = r
  Loop
  If Val(a + b * m) = p Then
  b = b
  a = a + b * (m - 1)
  d = d
  c = c + d * (m - 1)
  Else
  If Val(b + a * m) = p Then
  a = a
  b = b + a * m
  c = c
  d = d + c * m
  Else
  b = b
  a = a + b * (m - 1)
  d = d
  c = c + d * (m - 1)
  End If
  End If

 x = (a + b) Mod p
  Y = (c + d) Mod n
  
  
  Else
  
  If MBJC(Trim(n), Trim(p)) >= 1 Then
  m = n
  q = p
  s1 = 1
  Else
  m = p
  q = n
  s1 = 0
  End If
  Do Until zhengchuqyushu(MCC1(Trim(m), Trim(q))) = 0
  s = zhengchuqy(MCC1(Trim(m), Trim(q)))
  r = zhengchuqyushu(MCC1(Trim(m), Trim(q)))
  s1 = s1 + 1
  If s1 Mod 2 = 1 Then
  a = a
  b = MPC1(MbC(Trim(a), Trim(s)), Trim(b))
  c = c
  d = MPC1(MbC(Trim(c), Trim(s)), Trim(d))
  Else
  b = b
  a = MPC1(Trim(a), MbC(Trim(b), Trim(s)))
  d = d
  c = MPC1(Trim(c), MbC(Trim(d), Trim(s)))
  End If
  
  m = q
  q = r
  Loop
  
  If MPC1(Trim(a), MbC(Trim(b), Trim(m))) = p Then
  b = b
  a = MPC1(Trim(a), MbC(Trim(b), MPC(Trim(m), 1)))
  d = d
  c = MPC1(Trim(c), MbC(Trim(d), MPC(Trim(m), 1)))
  Else
  If MPC1(Trim(b), MbC(Trim(a), Trim(m))) = p Then
  a = a
  b = MPC1(Trim(b), MbC(Trim(a), Trim(m)))
  c = c
  d = MPC1(Trim(d), MbC(Trim(c), Trim(m)))
  Else
  b = b
  a = MPC1(Trim(a), MbC(Trim(b), MPC(Trim(m), 1)))
  d = d
  c = MPC1(Trim(c), MbC(Trim(d), MPC(Trim(m), 1)))
  End If
  End If
Do While Left(a, 1) = "0"
    a = Mid(a, 2)
Loop
  
  End If
  
  qniyuan = a

 End Function

 

 Private Function qksmimo(sa As String, sb As String, sc As String) As String

 Dim c, e, n, d
c = Trim(sa)
e = Trim(sb)
n = Trim(sc)
d = 1
If Len(c) < 5 And Len(e) < 5 And Len(n) < 5 Then
c = Val(c): n = Val(n)
Do While e > 0
If Right(e, 1) Mod 2 = 0 Then
c = c * c Mod n
e = e / 2

Else
d = d * c Mod n
e = e - 1
End If
Loop
Else
c = c
Do While MBJC(Trim(e), 1) >= 0
If Right(e, 1) Mod 2 = 0 Then
c = zhengchuqyushu(MCC1(MbC(Trim(c), Trim(c)), Trim(n)))
e = zhengchuqy(MCC1(Trim(e), 2))
Else
d = zhengchuqyushu(MCC1(MbC(Trim(c), Trim(d)), Trim(n)))
e = MPC(Trim(e), 1)
End If
Loop
End If

qksmimo = d

 End Function

 

 Private Function fenjieyinzi(sa As String) As String
Dim x, a, b
x = sa
b = Int(Sqr(Val(x)) / 2)
If x = 3 Or x = 2 Then
a = True
Else
If x Mod 2 = 0 Then
a = False
Else

For i = 3 To 2 * b + 1 Step 2
If x Mod i = 0 Then
a = False
Exit For

Else: a = True

End If
Next
End If
End If
If a = True Then
fenjieyinzi = "这是素数"
Else
fenjieyinzi = "2*2"
End If

End Function


Private Sub Command1_Click()
Dim a, n
Dim t As Double
t = Timer
n = Trim(Text1)
If Len(n) < 6 Then
Text2 = fenjieyinzi(Trim(n))
Else
n1 = MPC(Trim(n), 1)
a = 123
'a为明文
a1 = zzxc(Trim(n), Trim(a))
If Val(a1) > 1 Then
Text2 = a1 & "*"
Else
c = 999
'c为公钥
Do While zzxc(Trim(n1), Trim(c)) > 1
c = Val(c - 1)
Loop
d = qniyuan(Trim(c), Trim(n1))
'd为逆元为私钥
a2 = qksmimo(Trim(a), Trim(c), Trim(n))
'a2为密文
a3 = qksmimo(Trim(a2), Trim(d), Trim(n))
If MBJC(Trim(a3), Trim(a)) = 0 Then
Text2 = "这是素数有" & Len(n) & "位,用时" & Timer - t & "秒"
Else
Text2 = "2*2" & "用时" & Timer - t & "秒"
End If
End If
End If
End Sub

Private Sub Command2_Click()
  Text1 = "233333333333333333333333333333333333333333333333333333"
  Text2 = ""
End Sub

Private Sub Form_Load()
  Command2_Click
End Sub

[此贴子已经被作者于2020-2-22 09:02编辑过]

谢谢你！原理是成立的，我已经推导证明过了，你说的道理是对的。命题是可以逆推的，结论和前提是充要条件。所以是确定性的，比AKS法方便容易实现，比拉宾-米勒法靠谱，拉宾-米勒法在小数据中由于人们已经找到了漏掉的合数，所以可以排除确定，更大范围就不是确定的。这个法是确定的，限于速度，我也不知道能算到多少位的。
程序速度提高了就是有价值的，谢谢你，非常感谢！明天我试试！
前面提到的那个大整数的除法的算法我已经搞出程序，不快，前面的迭代增长太快，是2的幂数增长的，最后一步弄不准了，只好一个一个加上，所以最后这一步就特别慢，偶尔会有一个刚好不用加的就快，不好用。不是好法，不是普遍的所有整数的速度都能提高的。不发了。仅供参考。

这个文件打不开，我的电脑没有打开此文件的程序需要下载其他程序，也不知道啥样的程序。
谢谢您！
下面发个除法程序，只发主程序，也是个可调用程序。太慢，不断调用乘法比较大小，加法减法，若把乘法等程序速度提高了，程序是不是能用也不知道，发一下仅供参考。
 Public Function MCC1(D1 As String, D2 As String) As String ';大整数的除法
Dim ss
 ss = MBJC(D1, D2)
If ss = -1 Then
 MCC1 = "0" & "/" & D1
  Else
  If ss = 0 Then
   MCC1 = 1
   Else
   If Len(D2) < 9 Then
   MCC1 = MCC(Trim(D1), Trim(D2))
   Else
   x = Len(D1) - Len(D2)
   d4 = D1
  If x <= 3 Then
  x = 3
  B = 0
  Else
  a = Log(2) / Log(10)
  X1 = x
 B = Int(X1 * Val(a))
Do While B >= x
X1 = X1 + 1
B = Int(X1 * Val(a))
Loop
B = X1 - 1
End If
  c = 1
   Do Until i = B
    i = i + 1
    c = MbC(Trim(c), 2)
   Loop
   i1 = B
  Do While MBJC(Trim(D1), MbC(Trim(D2), Trim(c))) = 1
 i1 = i1 + 1
c = MbC(Trim(c), 2)
Loop
c = 1
Do Until i2 = i1 - 1
c = MbC(Trim(c), 2)
i2 = i2 + 1
Loop
D1 = MPC(Trim(D1), MbC(Trim(D2), Trim(c)))
D3 = c
C1 = c

D1 = MPC(Trim(d4), MbC(Trim(D3), Trim(D2)))
d5 = D1
If x >= 15 Then
x = x + 2
Else
x = x
End If
Do Until i2 <= x
c = 1
i1 = 0
  Do While MBJC(Trim(D1), MbC(Trim(D2), Trim(c))) = 1
 i1 = i1 + 1
c = MbC(Trim(c), 2)
Loop

i2 = 0
c = 1
Do Until i2 = i1 - 1
c = MbC(Trim(c), 2)
i2 = i2 + 1
Loop
D1 = MPC(Trim(D1), MbC(Trim(D2), Trim(c)))
D3 = MPC1(Trim(D3), Trim(c))
C1 = c
Print i2, D3
Loop
Do While MBJC(Trim(d4), MbC(Trim(D3), Trim(D2))) > 0
D3 = MPC1(Trim(D3), 1)
Loop
If MBJC(Trim(d4), MbC(Trim(D3), Trim(D2))) < 0 Then
D3 = MPC(Trim(D3), 1)
Else
D3 = D3
End If
JW = MPC(Trim(d4), MbC(Trim(D3), Trim(D2)))

If MBJC(Trim(JW), 0) = 0 Then
MCC1 = D3
Else

MCC1 = D3 & "/" & JW
End If
 End If
End If
End If
End Function