Answers:
Convert.ToString
可用于将数字转换为指定基数的等效字符串表示形式。
例:
string binary = Convert.ToString(5, 2); // convert 5 to its binary representation
Console.WriteLine(binary); // prints 101
但是,正如评论所指出的那样,Convert.ToString
仅支持以下有限的(但通常是足够的)基础集:2、8、10或16。
我不知道BCL中能够将数字转换为任何基数的任何方法,因此您必须编写自己的小型实用程序函数。一个简单的示例如下所示(请注意,通过替换字符串串联肯定可以使速度更快):
class Program
{
static void Main(string[] args)
{
// convert to binary
string binary = IntToString(42, new char[] { '0', '1' });
// convert to hexadecimal
string hex = IntToString(42,
new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
'A', 'B', 'C', 'D', 'E', 'F'});
// convert to hexavigesimal (base 26, A-Z)
string hexavigesimal = IntToString(42,
Enumerable.Range('A', 26).Select(x => (char)x).ToArray());
// convert to sexagesimal
string xx = IntToString(42,
new char[] { '0','1','2','3','4','5','6','7','8','9',
'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x'});
}
public static string IntToString(int value, char[] baseChars)
{
string result = string.Empty;
int targetBase = baseChars.Length;
do
{
result = baseChars[value % targetBase] + result;
value = value / targetBase;
}
while (value > 0);
return result;
}
/// <summary>
/// An optimized method using an array as buffer instead of
/// string concatenation. This is faster for return values having
/// a length > 1.
/// </summary>
public static string IntToStringFast(int value, char[] baseChars)
{
// 32 is the worst cast buffer size for base 2 and int.MaxValue
int i = 32;
char[] buffer = new char[i];
int targetBase= baseChars.Length;
do
{
buffer[--i] = baseChars[value % targetBase];
value = value / targetBase;
}
while (value > 0);
char[] result = new char[32 - i];
Array.Copy(buffer, i, result, 0, 32 - i);
return new string(result);
}
}
使用数组缓冲区而不是字符串连接来构建结果字符串可以改善性能,尤其是在数量较大时(请参见method IntToStringFast
)。在最佳情况下(即可能的最长输入),此方法的速度大约快三倍。但是,对于1位数字(即目标库中的1位数字),IntToString
速度会更快。
我最近在博客上写了这个。我的实现在计算过程中不使用任何字符串操作,因此非常快。支持转换为以2到36为底的任何数字系统:
/// <summary>
/// Converts the given decimal number to the numeral system with the
/// specified radix (in the range [2, 36]).
/// </summary>
/// <param name="decimalNumber">The number to convert.</param>
/// <param name="radix">The radix of the destination numeral system (in the range [2, 36]).</param>
/// <returns></returns>
public static string DecimalToArbitrarySystem(long decimalNumber, int radix)
{
const int BitsInLong = 64;
const string Digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
if (radix < 2 || radix > Digits.Length)
throw new ArgumentException("The radix must be >= 2 and <= " + Digits.Length.ToString());
if (decimalNumber == 0)
return "0";
int index = BitsInLong - 1;
long currentNumber = Math.Abs(decimalNumber);
char[] charArray = new char[BitsInLong];
while (currentNumber != 0)
{
int remainder = (int)(currentNumber % radix);
charArray[index--] = Digits[remainder];
currentNumber = currentNumber / radix;
}
string result = new String(charArray, index + 1, BitsInLong - index - 1);
if (decimalNumber < 0)
{
result = "-" + result;
}
return result;
}
万一有人需要,我还实现了快速逆函数: 任意到十进制数字系统。
result = "-" + result
什么 那是某种填充吗?如何修改代码,以便仅将AZ或0-9用于填充字符?
"-"
在result = "-" + result
看台上为负数的负号。这不是填充字符。
快速的“ 从 ”和“ 到 ”方法
我参加聚会很晚,但我对以前的答案进行了改进,并对其进行了改进。我认为这两种方法比迄今为止发布的任何其他方法都快。我能够在一台单核计算机上在不到400毫秒的时间内将1,000,000个数字从基数36转换为基数36。
下面的示例适用于基数62。更改BaseChars
数组以与其他任何基数进行相互转换。
private static readonly char[] BaseChars =
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".ToCharArray();
private static readonly Dictionary<char, int> CharValues = BaseChars
.Select((c,i)=>new {Char=c, Index=i})
.ToDictionary(c=>c.Char,c=>c.Index);
public static string LongToBase(long value)
{
long targetBase = BaseChars.Length;
// Determine exact number of characters to use.
char[] buffer = new char[Math.Max(
(int) Math.Ceiling(Math.Log(value + 1, targetBase)), 1)];
var i = buffer.Length;
do
{
buffer[--i] = BaseChars[value % targetBase];
value = value / targetBase;
}
while (value > 0);
return new string(buffer, i, buffer.Length - i);
}
public static long BaseToLong(string number)
{
char[] chrs = number.ToCharArray();
int m = chrs.Length - 1;
int n = BaseChars.Length, x;
long result = 0;
for (int i = 0; i < chrs.Length; i++)
{
x = CharValues[ chrs[i] ];
result += x * (long)Math.Pow(n, m--);
}
return result;
}
编辑(2018-07-12)
修复了解决@AdrianBotor(请参见注释)将46655转换为基数36时遇到的特殊情况的问题。这是由小的浮点错误计算引起的,该错误Math.Log(46656, 36)
恰好是3,但.NET返回3 + 4.44e-16
,这会在输出缓冲区中引起额外的字符。
BaseToLong(LongToBase(46655)) == 46655
0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ
并转换value 46655
。结果应该是,ZZZ
但是在调试器中我得到了\0ZZZ
。仅此值获得额外的\0
。例如,值46654
正确转换为ZZY
。
LongToBase
至return new string(buffer, (int) i, buffer.Length - (int)i);
参加这个聚会很晚,但是我最近为一个正在工作的项目编写了以下帮助程序类。它旨在将短字符串转换为数字并再次返回(这是一种简单的完美哈希函数),但是它还将在任意基数之间执行数字转换。该Base10ToString
方法实施回答了最初发布的问题。
需要使用shouldSupportRoundTripping
传递给类构造函数的标志,以防止在转换为以10为基数并再次转换时丢失数字字符串中的前导数字(根据我的要求,这是至关重要的!)。在大多数情况下,数字字符串中前导0的丢失可能不会成为问题。
无论如何,这是代码:
using System;
using System.Collections.Generic;
using System.Linq;
namespace StackOverflow
{
/// <summary>
/// Contains methods used to convert numbers between base-10 and another numbering system.
/// </summary>
/// <remarks>
/// <para>
/// This conversion class makes use of a set of characters that represent the digits used by the target
/// numbering system. For example, binary would use the digits 0 and 1, whereas hex would use the digits
/// 0 through 9 plus A through F. The digits do not have to be numerals.
/// </para>
/// <para>
/// The first digit in the sequence has special significance. If the number passed to the
/// <see cref="StringToBase10"/> method has leading digits that match the first digit, then those leading
/// digits will effectively be 'lost' during conversion. Much of the time this won't matter. For example,
/// "0F" hex will be converted to 15 decimal, but when converted back to hex it will become simply "F",
/// losing the leading "0". However, if the set of digits was A through Z, and the number "ABC" was
/// converted to base-10 and back again, then the leading "A" would be lost. The <see cref="System.Boolean"/>
/// flag passed to the constructor allows 'round-tripping' behaviour to be supported, which will prevent
/// leading digits from being lost during conversion.
/// </para>
/// <para>
/// Note that numeric overflow is probable when using longer strings and larger digit sets.
/// </para>
/// </remarks>
public class Base10Converter
{
const char NullDigit = '\0';
public Base10Converter(string digits, bool shouldSupportRoundTripping = false)
: this(digits.ToCharArray(), shouldSupportRoundTripping)
{
}
public Base10Converter(IEnumerable<char> digits, bool shouldSupportRoundTripping = false)
{
if (digits == null)
{
throw new ArgumentNullException("digits");
}
if (digits.Count() == 0)
{
throw new ArgumentException(
message: "The sequence is empty.",
paramName: "digits"
);
}
if (!digits.Distinct().SequenceEqual(digits))
{
throw new ArgumentException(
message: "There are duplicate characters in the sequence.",
paramName: "digits"
);
}
if (shouldSupportRoundTripping)
{
digits = (new[] { NullDigit }).Concat(digits);
}
_digitToIndexMap =
digits
.Select((digit, index) => new { digit, index })
.ToDictionary(keySelector: x => x.digit, elementSelector: x => x.index);
_radix = _digitToIndexMap.Count;
_indexToDigitMap =
_digitToIndexMap
.ToDictionary(keySelector: x => x.Value, elementSelector: x => x.Key);
}
readonly Dictionary<char, int> _digitToIndexMap;
readonly Dictionary<int, char> _indexToDigitMap;
readonly int _radix;
public long StringToBase10(string number)
{
Func<char, int, long> selector =
(c, i) =>
{
int power = number.Length - i - 1;
int digitIndex;
if (!_digitToIndexMap.TryGetValue(c, out digitIndex))
{
throw new ArgumentException(
message: String.Format("Number contains an invalid digit '{0}' at position {1}.", c, i),
paramName: "number"
);
}
return Convert.ToInt64(digitIndex * Math.Pow(_radix, power));
};
return number.Select(selector).Sum();
}
public string Base10ToString(long number)
{
if (number < 0)
{
throw new ArgumentOutOfRangeException(
message: "Value cannot be negative.",
paramName: "number"
);
}
string text = string.Empty;
long remainder;
do
{
number = Math.DivRem(number, _radix, out remainder);
char digit;
if (!_indexToDigitMap.TryGetValue((int) remainder, out digit) || digit == NullDigit)
{
throw new ArgumentException(
message: "Value cannot be converted given the set of digits used by this converter.",
paramName: "number"
);
}
text = digit + text;
}
while (number > 0);
return text;
}
}
}
也可以将其子类化以派生自定义数字转换器:
namespace StackOverflow
{
public sealed class BinaryNumberConverter : Base10Converter
{
public BinaryNumberConverter()
: base(digits: "01", shouldSupportRoundTripping: false)
{
}
}
public sealed class HexNumberConverter : Base10Converter
{
public HexNumberConverter()
: base(digits: "0123456789ABCDEF", shouldSupportRoundTripping: false)
{
}
}
}
代码将像这样使用:
using System.Diagnostics;
namespace StackOverflow
{
class Program
{
static void Main(string[] args)
{
{
var converter = new Base10Converter(
digits: "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz",
shouldSupportRoundTripping: true
);
long number = converter.StringToBase10("Atoz");
string text = converter.Base10ToString(number);
Debug.Assert(text == "Atoz");
}
{
var converter = new HexNumberConverter();
string text = converter.Base10ToString(255);
long number = converter.StringToBase10(text);
Debug.Assert(number == 255);
}
}
}
}
本论坛帖子中的此类课程对您有帮助吗?
public class BaseConverter {
public static string ToBase(string number, int start_base, int target_base) {
int base10 = this.ToBase10(number, start_base);
string rtn = this.FromBase10(base10, target_base);
return rtn;
}
public static int ToBase10(string number, int start_base) {
if (start_base < 2 || start_base > 36) return 0;
if (start_base == 10) return Convert.ToInt32(number);
char[] chrs = number.ToCharArray();
int m = chrs.Length - 1;
int n = start_base;
int x;
int rtn = 0;
foreach(char c in chrs) {
if (char.IsNumber(c))
x = int.Parse(c.ToString());
else
x = Convert.ToInt32(c) - 55;
rtn += x * (Convert.ToInt32(Math.Pow(n, m)));
m--;
}
return rtn;
}
public static string FromBase10(int number, int target_base) {
if (target_base < 2 || target_base > 36) return "";
if (target_base == 10) return number.ToString();
int n = target_base;
int q = number;
int r;
string rtn = "";
while (q >= n) {
r = q % n;
q = q / n;
if (r < 10)
rtn = r.ToString() + rtn;
else
rtn = Convert.ToChar(r + 55).ToString() + rtn;
}
if (q < 10)
rtn = q.ToString() + rtn;
else
rtn = Convert.ToChar(q + 55).ToString() + rtn;
return rtn;
}
}
完全未经测试...让我知道它是否有效!(复制粘贴,以防论坛帖子消失或其他问题。)
我也一直在寻找一种快速的方法来将十进制数转换为[2..36]范围内的另一个基数,因此我开发了以下代码。它很容易遵循,并使用Stringbuilder对象作为字符缓冲区的代理,我们可以按字符对字符进行索引。与替代方法相比,该代码看起来非常快,并且比初始化字符数组中的各个字符要快得多。
对于您自己的用途,您可能希望:1 /返回一个空字符串,而不是引发异常。2 /删除基数检查以使方法运行更快3 /用32'0初始化Stringbuilder对象并删除行结果.Remove(0,i);。这将导致字符串以前导零返回,并进一步提高了速度。4 /将Stringbuilder对象设置为类中的静态字段,因此无论调用DecimalToBase方法多少次,Stringbuilder对象仅初始化一次。如果执行此操作,则上面的更改3将不再起作用。
我希望有人觉得这很有用:)
原子悖论
static string DecimalToBase(int number, int radix)
{
// Check that the radix is between 2 and 36 inclusive
if ( radix < 2 || radix > 36 )
throw new ArgumentException("ConvertToBase(int number, int radix) - Radix must be between 2 and 36.");
// Create a buffer large enough to hold the largest int value represented in binary digits
StringBuilder result = new StringBuilder(" "); // 32 spaces
// The base conversion calculates the digits in reverse order so use
// an index to point to the last unused space in our buffer
int i = 32;
// Convert the number to the new base
do
{
int remainder = number % radix;
number = number / radix;
if(remainder <= 9)
result[--i] = (char)(remainder + '0'); // Converts [0..9] to ASCII ['0'..'9']
else
result[--i] = (char)(remainder + '7'); // Converts [10..36] to ASCII ['A'..'Z']
} while ( number > 0 );
// Remove the unwanted padding from the front of our buffer and return the result
// Note i points to the last unused character in our buffer
result.Remove( 0, i );
return (result.ToString());
}
我使用它来将Guid存储为较短的字符串(但仅限于使用106个字符)。如果有人感兴趣,这里是我的代码,用于将字符串解码回数字值(在这种情况下,我使用2个ulong作为Guid值,而不是对Int128进行编码(因为我使用的是3.5而不是4.0)。具有106个唯一字符的字符串const。ConvertLongsToBytes非常令人兴奋。
private static Guid B106ToGuid(string pStr)
{
try
{
ulong tMutl = 1, tL1 = 0, tL2 = 0, targetBase = (ulong)CODE.Length;
for (int i = 0; i < pStr.Length / 2; i++)
{
tL1 += (ulong)CODE.IndexOf(pStr[i]) * tMutl;
tL2 += (ulong)CODE.IndexOf(pStr[pStr.Length / 2 + i]) * tMutl;
tMutl *= targetBase;
}
return new Guid(ConvertLongsToBytes(tL1, tL2));
}
catch (Exception ex)
{
throw new Exception("B106ToGuid failed to convert string to Guid", ex);
}
}
除了有需要对“数字”进行数学运算外,我也有类似的需求。我在这里采纳了一些建议,并创建了一个可以完成所有这些有趣工作的类。它允许将任何unicode字符用于表示数字,并且也可以使用小数。
该类非常易于使用。只需将数字创建为New BaseNumber
,设置一些属性,然后关闭即可。例程负责自动在以10为基数和以x为基数之间进行切换,并且将您设置的值保留在您以它为基础的基数中,因此不会丢失任何精度(直到转换为止,但是即使这样精度损失也应该非常小)日常使用Double
以及Long
在任何可能的情况下)。
我无法命令此例程的速度。它可能很慢,所以我不确定它是否适合提出该问题的人的需求,但是它肯定是灵活的,因此希望其他人也可以使用它。
对于可能需要此代码来计算Excel中下一列的其他任何人,我都将包括我使用的利用此类的循环代码。
Public Class BaseNumber
Private _CharacterArray As List(Of Char)
Private _BaseXNumber As String
Private _Base10Number As Double?
Private NumberBaseLow As Integer
Private NumberBaseHigh As Integer
Private DecimalSeparator As Char = System.Globalization.CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator
Private GroupSeparator As Char = System.Globalization.CultureInfo.CurrentCulture.NumberFormat.NumberGroupSeparator
Public Sub UseCapsLetters()
'http://unicodelookup.com
TrySetBaseSet(65, 90)
End Sub
Public Function GetCharacterArray() As List(Of Char)
Return _CharacterArray
End Function
Public Sub SetCharacterArray(CharacterArray As String)
_CharacterArray = New List(Of Char)
_CharacterArray.AddRange(CharacterArray.ToList)
TrySetBaseSet(_CharacterArray)
End Sub
Public Sub SetCharacterArray(CharacterArray As List(Of Char))
_CharacterArray = CharacterArray
TrySetBaseSet(_CharacterArray)
End Sub
Public Sub SetNumber(Value As String)
_BaseXNumber = Value
_Base10Number = Nothing
End Sub
Public Sub SetNumber(Value As Double)
_Base10Number = Value
_BaseXNumber = Nothing
End Sub
Public Function GetBaseXNumber() As String
If _BaseXNumber IsNot Nothing Then
Return _BaseXNumber
Else
Return ToBaseString()
End If
End Function
Public Function GetBase10Number() As Double
If _Base10Number IsNot Nothing Then
Return _Base10Number
Else
Return ToBase10()
End If
End Function
Private Sub TrySetBaseSet(Values As List(Of Char))
For Each value As Char In _BaseXNumber
If Not Values.Contains(value) Then
Throw New ArgumentOutOfRangeException("The string has a value, " & value & ", not contained in the selected 'base' set.")
_CharacterArray.Clear()
DetermineNumberBase()
End If
Next
_CharacterArray = Values
End Sub
Private Sub TrySetBaseSet(LowValue As Integer, HighValue As Integer)
Dim HighLow As KeyValuePair(Of Integer, Integer) = GetHighLow()
If HighLow.Key < LowValue OrElse HighLow.Value > HighValue Then
Throw New ArgumentOutOfRangeException("The string has a value not contained in the selected 'base' set.")
_CharacterArray.Clear()
DetermineNumberBase()
End If
NumberBaseLow = LowValue
NumberBaseHigh = HighValue
End Sub
Private Function GetHighLow(Optional Values As List(Of Char) = Nothing) As KeyValuePair(Of Integer, Integer)
If Values Is Nothing Then
Values = _BaseXNumber.ToList
End If
Dim lowestValue As Integer = Convert.ToInt32(Values(0))
Dim highestValue As Integer = Convert.ToInt32(Values(0))
Dim currentValue As Integer
For Each value As Char In Values
If value <> DecimalSeparator AndAlso value <> GroupSeparator Then
currentValue = Convert.ToInt32(value)
If currentValue > highestValue Then
highestValue = currentValue
End If
If currentValue < lowestValue Then
currentValue = lowestValue
End If
End If
Next
Return New KeyValuePair(Of Integer, Integer)(lowestValue, highestValue)
End Function
Public Sub New(BaseXNumber As String)
_BaseXNumber = BaseXNumber
DetermineNumberBase()
End Sub
Public Sub New(BaseXNumber As String, NumberBase As Integer)
Me.New(BaseXNumber, Convert.ToInt32("0"c), NumberBase)
End Sub
Public Sub New(BaseXNumber As String, NumberBaseLow As Integer, NumberBaseHigh As Integer)
_BaseXNumber = BaseXNumber
Me.NumberBaseLow = NumberBaseLow
Me.NumberBaseHigh = NumberBaseHigh
End Sub
Public Sub New(Base10Number As Double)
_Base10Number = Base10Number
End Sub
Private Sub DetermineNumberBase()
Dim highestValue As Integer
Dim currentValue As Integer
For Each value As Char In _BaseXNumber
currentValue = Convert.ToInt32(value)
If currentValue > highestValue Then
highestValue = currentValue
End If
Next
NumberBaseHigh = highestValue
NumberBaseLow = Convert.ToInt32("0"c) 'assume 0 is the lowest
End Sub
Private Function ToBaseString() As String
Dim Base10Number As Double = _Base10Number
Dim intPart As Long = Math.Truncate(Base10Number)
Dim fracPart As Long = (Base10Number - intPart).ToString.Replace(DecimalSeparator, "")
Dim intPartString As String = ConvertIntToString(intPart)
Dim fracPartString As String = If(fracPart <> 0, DecimalSeparator & ConvertIntToString(fracPart), "")
Return intPartString & fracPartString
End Function
Private Function ToBase10() As Double
Dim intPartString As String = _BaseXNumber.Split(DecimalSeparator)(0).Replace(GroupSeparator, "")
Dim fracPartString As String = If(_BaseXNumber.Contains(DecimalSeparator), _BaseXNumber.Split(DecimalSeparator)(1), "")
Dim intPart As Long = ConvertStringToInt(intPartString)
Dim fracPartNumerator As Long = ConvertStringToInt(fracPartString)
Dim fracPartDenominator As Long = ConvertStringToInt(GetEncodedChar(1) & String.Join("", Enumerable.Repeat(GetEncodedChar(0), fracPartString.ToString.Length)))
Return Convert.ToDouble(intPart + fracPartNumerator / fracPartDenominator)
End Function
Private Function ConvertIntToString(ValueToConvert As Long) As String
Dim result As String = String.Empty
Dim targetBase As Long = GetEncodingCharsLength()
Do
result = GetEncodedChar(ValueToConvert Mod targetBase) & result
ValueToConvert = ValueToConvert \ targetBase
Loop While ValueToConvert > 0
Return result
End Function
Private Function ConvertStringToInt(ValueToConvert As String) As Long
Dim result As Long
Dim targetBase As Integer = GetEncodingCharsLength()
Dim startBase As Integer = GetEncodingCharsStartBase()
Dim value As Char
For x As Integer = 0 To ValueToConvert.Length - 1
value = ValueToConvert(x)
result += GetDecodedChar(value) * Convert.ToInt32(Math.Pow(GetEncodingCharsLength, ValueToConvert.Length - (x + 1)))
Next
Return result
End Function
Private Function GetEncodedChar(index As Integer) As Char
If _CharacterArray IsNot Nothing AndAlso _CharacterArray.Count > 0 Then
Return _CharacterArray(index)
Else
Return Convert.ToChar(index + NumberBaseLow)
End If
End Function
Private Function GetDecodedChar(character As Char) As Integer
If _CharacterArray IsNot Nothing AndAlso _CharacterArray.Count > 0 Then
Return _CharacterArray.IndexOf(character)
Else
Return Convert.ToInt32(character) - NumberBaseLow
End If
End Function
Private Function GetEncodingCharsLength() As Integer
If _CharacterArray IsNot Nothing AndAlso _CharacterArray.Count > 0 Then
Return _CharacterArray.Count
Else
Return NumberBaseHigh - NumberBaseLow + 1
End If
End Function
Private Function GetEncodingCharsStartBase() As Integer
If _CharacterArray IsNot Nothing AndAlso _CharacterArray.Count > 0 Then
Return GetHighLow.Key
Else
Return NumberBaseLow
End If
End Function
End Class
现在,让代码循环通过Excel列:
Public Function GetColumnList(DataSheetID As String) As List(Of String)
Dim workingColumn As New BaseNumber("A")
workingColumn.SetCharacterArray("@ABCDEFGHIJKLMNOPQRSTUVWXYZ")
Dim listOfPopulatedColumns As New List(Of String)
Dim countOfEmptyColumns As Integer
Dim colHasData As Boolean
Dim cellHasData As Boolean
Do
colHasData = True
cellHasData = False
For r As Integer = 1 To GetMaxRow(DataSheetID)
cellHasData = cellHasData Or XLGetCellValue(DataSheetID, workingColumn.GetBaseXNumber & r) <> ""
Next
colHasData = colHasData And cellHasData
'keep trying until we get 4 empty columns in a row
If colHasData Then
listOfPopulatedColumns.Add(workingColumn.GetBaseXNumber)
countOfEmptyColumns = 0
Else
countOfEmptyColumns += 1
End If
'we are already starting with column A, so increment after we check column A
Do
workingColumn.SetNumber(workingColumn.GetBase10Number + 1)
Loop Until Not workingColumn.GetBaseXNumber.Contains("@")
Loop Until countOfEmptyColumns > 3
Return listOfPopulatedColumns
End Function
您会注意到Excel部分的重要部分是0在重新计算的数字中由@标识。所以我只过滤掉所有带有@的数字,然后得到正确的序列(A,B,C,...,Z,AA,AB,AC等)。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConvertToAnyBase
{
class Program
{
static void Main(string[] args)
{
var baseNumber = int.Parse(Console.ReadLine());
var number = int.Parse(Console.ReadLine());
string conversion = "";
while(number!=0)
{
conversion += Convert.ToString(number % baseNumber);
number = number / baseNumber;
}
var conversion2 = conversion.ToArray().Reverse();
Console.WriteLine(string.Join("", conversion2));
}
}
}
如果有人在寻找VB选项,则基于Pavel的答案:
Public Shared Function ToBase(base10 As Long, Optional baseChars As String = "0123456789ABCDEFGHIJKLMNOPQRTSUVWXYZ") As String
If baseChars.Length < 2 Then Throw New ArgumentException("baseChars must be at least 2 chars long")
If base10 = 0 Then Return baseChars(0)
Dim isNegative = base10 < 0
Dim radix = baseChars.Length
Dim index As Integer = 64 'because it's how long a string will be if the basechars are 2 long (binary)
Dim chars(index) As Char '65 chars, 64 from above plus one for sign if it's negative
base10 = Math.Abs(base10)
While base10 > 0
chars(index) = baseChars(base10 Mod radix)
base10 \= radix
index -= 1
End While
If isNegative Then
chars(index) = "-"c
index -= 1
End If
Return New String(chars, index + 1, UBound(chars) - index)
End Function
这是一种相当简单的方法,但是可能不是最快的方法。它非常强大,因为它是可组合的。
public static IEnumerable<int> ToBase(this int x, int b)
{
IEnumerable<int> ToBaseReverse()
{
if (x == 0)
{
yield return 0;
yield break;
}
int z = x;
while (z > 0)
{
yield return z % b;
z = z / b;
}
}
return ToBaseReverse().Reverse();
}
将此与这种简单的扩展方法结合起来,现在就可以得到任何基础:
public static string ToBase(this int number, string digits) =>
String.Concat(number.ToBase(digits.Length).Select(x => digits[x]));
可以这样使用:
var result = 23.ToBase("01");
var result2 = 23.ToBase("012X");
Console.WriteLine(result);
Console.WriteLine(result2);
输出为:
10111 11倍