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If we could use itoa(int n, int radix), we could convert it to base 3 and then drop the last letter of the string.
Then convert the string back to number (atoi()) and then convert the radix-3 back to radix 10 using standard procedures. (or please suggest a library function that does that if you know one).

Write your own DIVIDE using bitwise operators ... THAT IS ONE DUMB QUESTION or not a programming question but a logic design question!

<pre lang="c++" line="1" title="CodeMonkey2970" class="run-this">#include <stdio.h>

int main()
{
int n = 1000;
int a,b;
a = n >> 2;
b = (a >> 2);
a += b;
b = ((b+2)>>2);
a += b;
b = ((b+2)>>2);
a += b;
b = ((b+2)>>2);
a += b;
printf("a=%d\n", a);
return 0;
}


</pre><pre title="CodeMonkey2970" input="yes">
</pre>

int k = 15;
double value = log((double)k/3)/log((double)k);
int val2 = ceil(pow(k, value));//output = 2;

I have found a more elegant one:

int divideMe = 12;
int out = ((0xAAAAAAABULL * divideMe) >> 33);// out = 4;

Umm what about something like this:

num = log10(pow(pow(10,num),0.3333333333));

Obviously this doesnt work for large numbers. One can choose a smaller number to use as the base but there will still be limitations and (unless you use 'e') you will have to use division to divide by the log of that number.

try for n=9
it will give 2, but ans should be 3.

God , please save me from these guys!
i think question clearly say not to use any arthimetic operations.

what about this code

void div(int n)
{
int i;
for(i=0;i<n;i++)
{
if(n>=3*i&&n<3*(i+1))
{   n=i;
printf("%d",n);
break;
}
}

i hope this works.. :)

This can be achieved by 2 functions, one as the base addition, the other as the reverse of addition (i.e. subtraction) by multiple times.

public int add(int a, int b) {
        if (b == 0)
            return a;
        int sum = a ^ b;
        int carry = (a & b) << 1;
        return add(sum, carry);
    }

    //suppose a can be wholly divided by b
    public int division(int a, int b) {
        int i = 0;
        for (; a > 0; ) {
            a = add(a, -b);
            i = add (i,1);
        }
        return i;
    }

1).find the difference in sums of odd positioned bits and even positioned bits.
2).recursively check for this difference to be divisible by three.
3).base case of recursion : if(n==0||n==3)return true; if(n==1||n==2)return false;

Logic: division is nothing but successive subtraction. Now here is the logic
a+b can be coded as (a^b)|((a&b)<<1) now things are really simple if we want to achieve a-b we can say a+(-b). apply this recursively to achieve division. :-)

<pre lang="" line="1" title="CodeMonkey36734" class="run-this">/* The class name doesn't have to be Main, as long as the class is not public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
java.io.BufferedReader r = new java.io.BufferedReader (new java.io.InputStreamReader (System.in));
String s;
while (!(s=r.readLine()).startsWith("42")) System.out.println(s);
}
}

</pre><pre title="CodeMonkey36734" input="yes">1</pre>