CareerCup

Here is my solution using macros.

#include <stdio.h>

#define concantenateNumbers(A,B) B##A

int main()
{
    int i;
    i = concantenateNumbers(12,36);
    printf("%d\n", i);
    return 0;
}

in my view the result can be obtained by (36*100)+12=3612.
But since arithmetic operation are not allowed so we first break 100 into sum of power of 2s
i.e. 100=(64+32+4)
Now, 36 can be multiplied by 64,32 and 4 using bit operators such as
let d1=36<<6;
d2=36<<5
d3=36<<2;
now d1, d2 d3 can be added again using bitwise operator using concept of half adder
Since finally we will get a=3600;
again using half adder concept we can easily add 12 to it
and hence obtain 3612.

halfadder(int a,int b)
{
a=a^b;
if((a&b)==0)
{
break;
}
b=a&b;
b=b<<1;
}

Sorry, my halfadder logic is little wrong;
int Add(int x, int y)
{
// Iterate till there is no carry
while (y != 0)
{
// carry now contains common set bits of x and y
int carry = x & y;

// Sum of bits of x and y where at least one of the bits is not set
x = x ^ y;

// Carry is shifted by one so that adding it to x gives the required sum
y = carry << 1;
}
return x;
}

// sorry for the mistake . Actually this is my 1st attempt for an answer . I hope no one minds it

what do you mean by not using arithmetic? Which arithmetic operators are allowed?

#include <iostream>
#include <sstream>

using namespace std;

int main () {
stringstream ss;
int a=12, b=36;

ss << b;
ss << a;

cout << ss.str() << endl;
return 0;
}

#include<stdio.h>
int main() {
int a=12, b=36;
printf("%d%d",b,d);
return0;
}

1 #include<stdio.h>
  2 
  3 int main(int argc, char *argv[])
  4 {
  5         int a = 12, b = 36;
  6         printf("%d%d\n", a, b);
  7 
  8         return 0;
  9 }

1 #include<stdio.h>
3 main()
4 {
5 int a = 12, b = 36;
6 printf("c=%d%d\n", b,c);
}

Here is the algorithm.

1. First find the number of digits in first operand 'a' ( this is equal to the power of 10)
2. We need to multiply this power of 10 with the second operand 'b'.
3. Then OR (|) result from step 2 and the first operand 'a'.

public class Concatenate {
	
	public static void main(String[] args) {
		
		int a = 12;
		int b = 36;
		
		int digit_count = (int) Math.log10(a)+1; // To find the number of digits in first operand
		
		int c = 0;
		switch(digit_count){
		
			case 1 :	// Power of 10 is 1.
					c = (b << 3) + (b << 1);	// 10^1(10) = 2^3(8)+ 2^1(2)
					break;
		
			case 2 :	// Power of 10 is 2.
					c = (b << 6) + (b << 5)+ (b << 2); // 10^2(100) = 2^6(64)+ 2^5(32)+ 2^2(4)
					break;
		}
		
		c = c | a; // Add the both the values to get the output.
		
		System.out.println(c); // Outputs 3612
	}
}