如何在VB.Net中使用与biginteger相当的indexof（）？

I am trying to sum up the digits in a very large number. I have gotten the length of the number with
l = answer.bitLength() but I can't figure out how to increament through each digit using a For loop. Any ideas?

我试图总结一个非常大的数字。我已经得到了l = answer.bitLength()的数字长度,但我无法弄清楚如何使用For循环增加每个数字。有任何想法吗?

I'm using the java.math.biginteger.

我正在使用java.math.biginteger。

Visual Studio 2005 Version 2.0

Visual Studio 2005版本2.0

I should also add that I can't seem to use <> or any of the simple math options with the biginteger I'm using. If anyone could tell me how to use a different biginteger I would be more than willing to swap.

我还应该补充一点,我似乎无法使用<>或任何简单的数学选项与我正在使用的biginteger。如果有人能告诉我如何使用不同的大整数,我会更愿意交换。

Dim answer As java.math.BigInteger
Dim sum As Integer = 0
Dim x As Integer
Dim i As Integer
'Sets value of answer equal to 1
answer = java.math.BigInteger.valueOf(1)

'gets 100!
For i = 1 To 100
answer = answer.multiply(java.math.BigInteger.valueOf(i))
Next

'gets length of answer
Dim l As Integer
l = answer.bitLength()

'Sums up digits in 100!
For x = 0 To l - 1
'Need to pull each character here to add them all up
Next

Final Solution for summing up the digits. Thanks to wageoghe.

总结数字的最终解决方案。感谢payoghe。

Dim r As Integer
Dim s As Integer
s = 0
While (answer.compareTo(java.math.BigInteger.valueOf(0)) > 0)

r = answer.mod(java.math.BigInteger.valueOf(10)).ToString()
s = s + r
answer = answer.divide(java.math.BigInteger.valueOf(10))

End While

3 个解决方案

#1

Something like this should work:

像这样的东西应该工作:

    Dim bi As New System.Numerics.BigInteger(12345)
    Dim c As Char
    Dim s As Long

    s = 0

    For Each c In bi.ToString()
        s = s + Integer.Parse(c.ToString())
    Next

Or this more conventional way using Mod and / (integer division)

或者这种更传统的方式使用Mod和/(整数除法)

    Dim bi As New System.Numerics.BigInteger(12345)
    Dim s As Long
    Dim r As Integer

    s = 0

    While bi <> 0
        r = bi Mod 10
        s = s + r
        bi = bi / 10
    End While

#2

If you think of the number as a list of binary characters, then you could get the least significant hex digit by ANDing the number with 0xF. If you then shifted the number right by 4 bits (>> 4), then you could get the next hex digit.

如果您将数字视为二进制字符列表,那么您可以通过将数字与0xF进行AND运算来获得最低有效十六进制数字。如果你然后将数字右移4位(>> 4),那么你可以获得下一个十六进制数字。

After you get all of the hex digits, you could sum them up and then convert them to decimal.

获得所有十六进制数字后,您可以将它们相加,然后将它们转换为十进制数字。

#3

Another approach, is to do the following (this assumes that answer is positive):

另一种方法是执行以下操作(假设答案是肯定的):

int sum = 0;
while(answer > 0){
    sum += answer % 10;
    answer /= 10;
}

#1