    def increment(a):
        import urllib2
        url = "http://www.html2xml.nl/Services/Calculator/Version1/Calculator.asmx/Add?a=%d&b=1" % a       
        result = urllib2.urlopen(url).read() 
        from xml.dom.minidom import parseString
        result = parseString(result)
        return int(result.getElementsByTagName('int')[0].childNodes[0].data)

Re: Calculate n + 1 without using + or - or * or /

#14

Post by namank » Sun, Jul 03, 2011, 3:02 PM UTC

icebraining wrote: post #13

I have the proper Enterprise version: def increment(a): import urllib2 url = "http://www.html2xml.nl/Services/Calculator/Version1/Calculator.asmx/Add?a=%d&b=1" % a result = urllib2.urlopen(url).read() from xml.dom.minidom import parseString result = parseString(result) return int(result.getElementsByTagName('int')[0].childNodes[0].data)

I'm pretty sure that's not a valid solution. Its the same as using Math.sum(). While it evaluates your resourcefulness, the point of the question, I think, is to figure out your knowledge and creative thinking

Re: Calculate n + 1 without using + or - or * or /

#15

Post by custominstall » Mon, Jul 04, 2011, 3:59 AM UTC

kirillzubovsky wrote: post #4

awaage wrote: post #3

how about this (example in ruby): def plus_1(x); arr = []; arr[x] = true; return arr.size; end

Funny, this is the first thing I thought of, but then realized it would only work if x>0

lol .. but also wont work for decimals

Re: Calculate n + 1 without using + or - or * or /

#16

Post by custominstall » Mon, Jul 04, 2011, 4:44 AM UTC

php solution provided number is a positive int :

round($n.'.9')

Re: Calculate n + 1 without using + or - or * or /

#17

Post by anujkk » Mon, Jul 04, 2011, 7:41 AM UTC

If it is just about not using +,-,* or / operators, we can do it by using "++" operator.

x=n++;

Re: Calculate n + 1 without using + or - or * or /

#18

Post by alecbenzer » Mon, Jul 04, 2011, 9:52 PM UTC

if n is an integer, I believe this works:

https://gist.github.com/1063990

uses only bit shits, bitwise XOR, and bitwise AND

I would imagine they meant not to use the operations, not just the symbols.

Re: Calculate n + 1 without using + or - or * or /

#19

Post by veyron » Mon, Jul 04, 2011, 10:48 PM UTC

Basically replicate the carry process in the form of bit ops:

    def inc(n):
        x = 1
        while n & 1 == 1:
            n = n >> 1
            x = x > 1
            n = n


    
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