这是我用python编写的解决方案

import math

num = 3500
row_number = str(math.ceil(num / 702))
letters = ''
num = num - 702 * math.floor(num / 702)
while num:
    mod = (num - 1) % 26
    letters += chr(mod + 65)
    num = (num - 1) // 26
result = row_number + ("".join(reversed(letters)))
print(result)

另一个解决方案:

private void Foo()
{
   l_ExcelApp = new Excel.ApplicationClass();
   l_ExcelApp.ReferenceStyle = Excel.XlReferenceStyle.xlR1C1;
   // ... now reference by R[row]C[column], Ex. A1 <==> R1C1, C6 <==> R3C6, ...
}

在这里查看更多- Excel中的单元格引用!作者:Nitin Paranjape博士

int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol >= 26)
{
    int nChar = nCol % 26;
    nCol = (nCol - nChar) / 26;
    // You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
    sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;

更新:Peter的评论是正确的。这就是我在浏览器中编写代码的结果。:-)我的解决方案是不编译，它是最左边的字母，它是在反向顺序构建字符串-现在都固定了。

除了bug之外，该算法基本上是将一个数字从10进制转换为26进制。

更新2:Joel Coehoorn是对的-上面的代码将返回AB为27。如果它是一个以26为底的实数，AA就等于A, Z之后的下一个数字就是BA。

int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol > 26)
{
    int nChar = nCol % 26;
    if (nChar == 0)
        nChar = 26;
    nCol = (nCol - nChar) / 26;
    sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
    sCol = sChars[nCol] + sCol;

在查看了这里提供的所有版本后，我决定自己使用递归来做一个。

这是我的vb.net版本:

Function CL(ByVal x As Integer) As String
    If x >= 1 And x <= 26 Then
        CL = Chr(x + 64)
    Else
        CL = CL((x - x Mod 26) / 26) & Chr((x Mod 26) + 1 + 64)
    End If
End Function

如果有人需要在没有VBA的Excel中做到这一点，这里有一种方法:

=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")

其中colNum是列号

在VBA中:

Function GetColumnName(colNum As Integer) As String
    Dim d As Integer
    Dim m As Integer
    Dim name As String
    d = colNum
    name = ""
    Do While (d > 0)
        m = (d - 1) Mod 26
        name = Chr(65 + m) + name
        d = Int((d - m) / 26)
    Loop
    GetColumnName = name
End Function

Objective-C实现:

-(NSString*)getColumnName:(int)n {
     NSString *name = @"";
     while (n>0) {
     n--;
     char c = (char)('A' + n%26);
     name = [NSString stringWithFormat:@"%c%@",c,name];
     n = n/26;
  }    
     return name;

}

迅速实现:

func getColumnName(n:Int)->String{
 var columnName = ""
 var index = n
 while index>0 {
     index--
     let char = Character(UnicodeScalar(65 + index%26))
     columnName = "\(char)\(columnName)"
     index = index / 26
 }
 return columnName

}

答案是基于:https://stackoverflow.com/a/4532562/2231118