0 = A
1 = B
...
25 = Z
26 = AA
27 = AB
...
701 = ZZ
702 = AAA

I cannot think of any solution that does not involve loop-bruteforce :-(

I expect a function/program, that accepts a decimal number and returns a string as a result.

stusmith

Haskell, 78 57 50 43 chars

o=map(['A'..'Z']:)$[]:o
e=(!!)$o>>=sequence

Other entries aren't counting the driver, which adds another 40 chars:

main=interact$unlines.map(e.read).lines

A new approach, using a lazy, infinite list, and the power of Monads! And besides, using sequence makes me :), using infinite lists makes me :o

If you look carefully the excel representation is like base 26 number but not exactly same as base 26.

In Excel conversion Z + 1 = AA while in base-26 Z + 1 = BA

The algorithm is almost same as decimal to base-26 conversion with just once change. In base-26, we do a recursive call by passing it the quotient, but here we pass it quotient-1:

function decimalToExcel(num)

        // base condition of recursion.
        if num < 26
                print 'A' + num 

        else                     
                quotient = num / 26;
                reminder = num % 26;

                // recursive calls.
                decimalToExcel(quotient - 1);
                decimalToExcel(reminder);
        end-if                       
end-function

Java Implementation

Python, 44 chars

Oh c'mon, we can do better than lengths of 100+ :

X=lambda n:~n and X(n/26-1)+chr(65+n%26)or''

Testing:

>>> for i in 0, 1, 25, 26, 27, 700, 701, 702:
...     print i,'=',X(i)
...     
0 = A
1 = B
25 = Z
26 = AA
27 = AB
700 = ZY
701 = ZZ
702 = AAA

Since I am not sure what base you're converting from and what base you want (your title suggests one and your question the opposite), I'll cover both.

Algorithm for converting `ZZ` to `701`

First recognize that we have a number encoded in base 26, where the "digits" are A..Z. Set a counter a to zero and start reading the number at the rightmost (least significant digit). Progressing from right to left, read each number and convert its "digit" to a decimal number. Multiply this by 26^a and add this to the result. Increment a and process the next digit.

Algorithm for converting `701` to `ZZ`

We simply factor the number into powers of 26, much like we do when converting to binary. Simply take num%26, convert it to A..Z "digits" and append to the converted number (assuming it's a string), then integer-divide your number. Repeat until num is zero. After this, reverse the converted number string to have the most significant bit first.

Edit: As you point out, once two-digit numbers are reached we actually have base 27 for all non-least-significant bits. Simply apply the same algorithms here, incrementing any "constants" by one. Should work, but I haven't tried it myself.

Re-edit: For the ZZ->701 case, don't increment the base exponent. Do however keep in mind that A no longer is 0 (but 1) and so forth.

Explanation of why this is not a base 26 conversion

Let's start by looking at the real base 26 positional system. (Rather, look as base 4 since it's less numbers). The following is true (assuming A = 0):

 A = AA = A * 4^1 + A * 4^0 = 0 * 4^1 + 0 * 4^0 = 0
 B = AB = A * 4^1 + B * 4^0 = 0 * 4^1 + 1 * 4^0 = 1
 C = AC = A * 4^1 + C * 4^0 = 0 * 4^1 + 2 * 4^0 = 2
 D = AD = A * 4^1 + D * 4^0 = 0 * 4^1 + 3 * 4^0 = 3
     BA = B * 4^0 + A * 4^0 = 1 * 4^1 + 0 * 4^0 = 4

And so forth... notice that AA is 0 rather than 4 as it would be in Excel notation. Hence, Excel notation is not base 26.

In Excel VBA ... the obvious choice :)

Sub a()

  For Each O In Range("A1:AA1")
    k = O.Address()
    Debug.Print Mid(k, 2, Len(k) - 3); "="; O.Column - 1
  Next

End Sub

Or for getting the column number in the first row of the WorkSheet (which make more sense, since we are in Excel ...)

Sub a()

  For Each O In Range("A1:AA1")
    O.Value = O.Column - 1
  Next

End Sub

Or better yet:

56 chars

Sub a()
  Set O = Range("A1:AA1")
  O.Formula = "=Column()"
End Sub

Scala: 63 chars

def c(n:Int):String=(if(n<26)""else c(n/26-1))+(65+n%26).toChar

Prolog, 109 123 bytes

Convert from decimal number to Excel string:

c(D,E):- d(D,X),atom_codes(E,X).
d(D,[E]):-D<26,E is D+65,!.
d(D,[O|M]):-N is D//27,d(N,M),O is 65+D rem 26.

That code does not work for c(27, N), which yields N='BB'

This one works fine:

c(D,E):-c(D,26,[],X),atom_codes(E,X).
c(D,B,T,M):-(D<B->M-S=[O|T]-B;(S=26,N is D//S,c(N,27,[O|T],M))),O is 91-S+D rem B,!.

Tests:

?- c(0, N).
N = 'A'.


?- c(27, N).
N = 'AB'.

?- c(701, N).
N = 'ZZ'.

?- c(702, N).
N = 'AAA'

Converts from Excel string to decimal number (87 bytes):

x(E,D):-x(E,0,D).
x([C],X,N):-N is X+C-65,!.
x([C|T],X,N):-Y is (X+C-64)*26,x(T,Y,N).

F# : 166 137

let rec c x  = if x < 26 then [(char) ((int 'A') + x)] else List.append (c (x/26-1)) (c (x%26))
let s x = new string (c x |> List.toArray)

PHP: At least 59 and 33 characters.

<?for($a=NUM+1;$a>=1;$a=$a/26)$c=chr(--$a%26+65).$c;echo$c;

Or the shortest version:

<?for($a=A;$i++<NUM;++$a);echo$a;

Using the following formula, you can figure out the last character in the string:

transform(int num)
    return (char)num + 47; // Transform int to ascii alphabetic char. 47 might not be right.

char lastChar(int num)
{
    return transform(num % 26);
}

Using this, we can make a recursive function (I don't think its brute force).

string getExcelHeader(int decimal)
{
    if (decimal > 26)
        return getExcelHeader(decimal / 26) + transform(decimal % 26);
    else
        return transform(decimal);
}

Or.. something like that. I'm really tired, maybe I should stop answering questions and go to bed :P

来源：https://stackoverflow.com/questions/4447550/convert-decimal-number-to-excel-header-like-number

标签

language-agnostic

code-golf

Convert decimal number to excel-header-like number