CareerCup

int Mul(int a,int b)
{
	int ans=0;
	int sign= ((a^b)>>31);
	a=abs(a);
	b=abs(b);
	while(b)
	{
		if(b & 0x01 )ans += a;
		b >>= 1;
		a <<= 1;
	}
	return sign?(~ans+1):ans;
}

int multiply( int a, int b) {
  if (a==0 || b==0)
    return 0;
  int sign = 0, result;
  if ((a < 0 && b > 0) || (a > 0 && b < 0)) {
    sign = 1;
  }
  int a1 = abs(a);
  int b1 = abs(b);
  while (b1--) {
    result + = a1;
  }
  if (sign) {
    return result/(-1);
  } else {
    return result;
  }
}

public class Solution {
    public static int multiply(int a, int b) {
	if (a == 0 || b == 0) {
	    return 0;
	}
	if (b == 1) {
	    return a;
	}
	boolean positiveFlag = true;
	if ((a < 0 && b > 0) || (a > 0 && b < 0)) {
	    positiveFlag = false;
	}
	a = Math.abs(a);
	b = Math.abs(b);

	if (a < b) {
	    a ^= b;
	    b ^= a;
	    a ^= b;
	}
	
	int ret1 = multiply(a, b/2);
	int ret2 = multiply(a, b%2);
	if (positiveFlag == false) {
	    return -(ret1+ret1+ret2)
	}
	else {
	    return ret1+ret1+ret2;
	}
    }

}

int multiply(int a, int b)
{
if(b == 0)
return 0;
if(b > 0)
return a + multiply(a, b-1);
if(b < 0)
return (-a) + multiply(a, b+1);

}

ex: x*y
int RES = 0;
for(int i = 0; i<x;i++)
{
RES += y;
}

int multiply( int a, int b )
{
    if( a==0 || b==0 )
        return 0;
    return (a/(1.0/b));
}

public static int multiply(int a, int b){
if(a==0||b==0) return 0;
int sign = 0;
if(((a>>31) &0x1) == 1){
sign = 1;
a = ~a + 1;
}if(((b>>31) & 0x1) == 1){
if (sign == 0)sign = 1;
else sign = 0;
b = ~b + 1;
}

int mul = 0;
//System.out.println(b + " "+ a);
while(b>0){
if((b & 0x1) > 0){
mul +=a;
}
b = b>>1;
a = a<<1;
}
if (sign ==1)
return ~mul + 1 ;

return mul;
}

Lucky Fellow to get this question

int Multiply(int number, int multiple)
{
int num=number;
while(multiple--)
number=number+num;
retrun number;
}

/**
*
* PropertiesFor the natural numbers, integers, fractions, and real and complex numbers, multiplication has certain properties:

Commutative property : x*y=y*x;
The order in which two numbers are multiplied does not matter;
Associative property : (x*y)*z=x*(y*z);
Expressions solely involving multiplication are invariant with respect to order of operations;
Distributive property : x*(y+z) = x*y+x*z;
Holds with respect to multiplication over addition. This identity is of prime importance in simplifying algebraic expressions:
Identity element : x*1=x;
The multiplicative identity is 1; anything multiplied by one is itself. This is known as the identity property:
Zero element : x*0=0;
Any number multiplied by zero is zero. This is known as the zero property of multiplication:
Zero is sometimes not included amongst the natural numbers.
There are a number of further properties of multiplication not satisfied by all types of numbers.
Negation : x*-1 = -x;
Negative one times any number is equal to the opposite of that number.
-1*-1=1;
Negative one times negative one is positive one.
The natural numbers do not include negative numbers.
Inverse element
Every number x, except zero, has a multiplicative inverse, , such that .
The natural numbers and integers do not have inverses.
Exculding folating and complex numbers mulitple
*/
public static long mulitple (int a, int b){
long mulitplicationResult = 0;
boolean isPositiveResult = false;
// Retrun zero if any one input has zero
if (a == 0 || b==0){
return mulitplicationResult;
}
//Sign calculation of result
if( (a<0 && b<0) || (a>0 && b>0) ){
isPositiveResult = true;
}else{
isPositiveResult = false;
}
//Converting -ve values of the input into +ve
if(a<0) a=(~a+1);
if(b<0) b=(~b+1);
//For example: 3 multiplied by 4 can also be calculated by adding 3 copies of 4 together.
while( b>0 ){
mulitplicationResult += a;
b--;
}
return isPositiveResult?mulitplicationResult:(~mulitplicationResult+1);
}

suppose we have a , b
1) First produce arr[9] where arr[i]=i*a , (1<=i<=9). we can do it with 9 addition.
2) For each digit in b[i]=k add i zero on right-side of the arr[k] and add them to each other.
The result is a*b.
It is the obvious way that we calculate multiply on paper!
This solution is useful for bignumber multiplication too.

this is a solution in java and it doesn't involve a loop, the answer is pretty simple:

public class Multiply{

	public static void main (String[] args){
		System.out.println("The result is:" + multiply(5,-13));
	}

	private static int multiply(int i, int j) {
		double k = i;
		double l = j;		
		return (int)(k/(1/l));
	}
}

This solution works if multiplication is among either positive or negative numbers. The ~ operator is used just to make positive number to negative and negative number to positive

// multiply 2 no without using mul operat.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<stdio.h>
using namespace std;
void main()
{
	int a, b;
	int flag1=0,mul=0;
	int flag2=0;
	printf("enter the two numbers\n");
	scanf_s("%d%d",&a,&b);
	if(a<0)
	{
	flag1=1;
	a=~(a)+1;
	}
	if(b<0)
	{
	flag2=1;
	b=~(b)+1;
	}
	for(int i=0;i<b;i++)
	{
		mul=mul+a;
		printf("the mul inside for loop %d\n",mul);
	}
	if(flag1==1 || flag2==1)
	{
		if(flag1==1 && flag2==1)
		{
			printf("both numbers are negative\n");
		}
		else
		{
			mul=~(mul)+1;
		}
	}
	printf("%d",mul);
}

function multiply(a, b)
  return exp(ln(a) + ln(b)

Can any one help me to find the multiplication of two floating point numbers without using * operator or any library fn that uses * internally....
I think we can take the integral & fractional part separately and multiply like above mentioned code for integral part. But m having difficulty with fractional part ...But if any one can help me to identify the bit position of fractional part in float, then it can be done easily....
Thanks in advance...

int function multiply(int a, int b)
{
	int result = 0;
	if ( b >= 0 )
	{
		while( b-- > 0 )
			result += a;
	}
	else
	{
		while( b++ < 0 )
			result -= a;
	}

	return $result;
}

public int sum(int a, int b) {
		int result = a ^ b;
		int carry = a & b;
		while (carry != 0) {
			carry <<=  1;
			int x = result;
			result ^= carry;
			carry &= x;
		}
		return result;		
	}

	public int multiply(int a, int b) {
		int result = 0; 
		if (a == 0 || b == 0) {
			return result;
		}
		boolean negative = a > 0 && b > 0 || a < 0 && b < 0 ? false : true;
		a = a > 0 ? a : -a;
		b = b > 0 ? b : -b;

		int k = 0;
		while (b != 0) {
			if ((b & 1) > 0) {
			    result = sum(result, a << k);
			}
			k++;
			b = b >>> 1;
		}
		if (negative)
			result = sum(~result, 1);

		return result;
	}

method with a helper function

def multiply(a,b):                
    sign = (a<0 and b < 0) or (a>0 and b> 0)                        
    def multiplyHelper(a,b):
        if b == 0 or a == 0:
            return 0
        if b%2 == 0:
            return multiplyHelper(a, b/2) + multiplyHelper(a, b/2)
        else:
            return a + multiplyHelper(a, b-1)        
    ans = multiplyHelper(abs(a), abs(b))    
    return ans if sign else ~ans+1
                                    
assert 35 == multiply(-7,-5)
assert -35 == multiply(-7, 5)

private static int multiply(int a, int b) {
		
		if(a == 0 || b == 0)
			return 0;
		
		int res = 0;
		int absA = Math.abs(a);
		int absB = Math.abs(b);
		if ((a > 0 && b > 0)  || (a < 0 && b < 0)) {
			a = absA;
			b = absB;
		}
		if( absA > absB) {
			while(absB > 0) {
				res+=a;
				absB--;
			}
		} else {
			while(absA > 0) {
				res+=b;
				absA--;
			}
		}
		return res;
	}

Try something like this:

public int myMethod(int i, int j) {
int x = 0;
if (j >= 0) {
for (int k = 1; k <= j; k++) {
x = x + i;
}
return x;

} else if ((i < 0) && (j < 0)) {
for (int k = 1; k <= -i; k++) {
x = x + j;
}
return (-x);

} else {
for (int k = 1; k <= i; k++) {
x = x + j;
}
return x;

}

}

{
int main(){
int a,b;
int ad=0;
cout<<" ";
cin>>a>>b;
for(int i=0;i<b;i++){
ad=ad+a;
}
cout<<"multiplication is \n"<<ad<<endl;


getch();
return 0;
}
}

int multiply(int a,int b)
{
while(b!=0)
{ if(b&1)
{ result=result+a;}
a=a<<1;
b=b>>1;
}
return result;
}

def multi(a:Int, b:Int):Int ={ if(b==1) a else multi(a+a,b-1);}

The function should return long for two int.

long mulwithoutmul(int a, int b)
{
	long res=0;
	long la;
	int i;
	char neg = ((a^b)>>31);
	a = abs(a);
	b = abs(b);
	if(a == 0 || b == 0)
		return 0;
	la = a;
	for(i = 0 ; i < 31 ; i++)
	{
		if((b>>i)&0x00000001)
			res+=la<<i;
	}
	if(neg)
	{
		res = 0-res;
	}
	return res;
}

!python
def multiply(x, y):
    if x == 0 or y == 0: return 0

    negative = False
    if x < 0 and y < 0: x = -x; y = -y; 
    elif x < 0: x = -x; negative = True
    elif y < 0: y = -y; negative = True        

    # Small optimization
    add_value     = x
    counter_range = y
    if y > x: add_value = y; counter_range = x

    value = 0
    for i in range(counter_range):
        value = value + add_value

    return -value if negative else value

why not a recursive algo like this?

public static int simpleMultiply(int one, int two)
{
if(two==0 || one ==0)
return 0;
if(one>0 && two>0)
{
return one+simpleMultiply(one,two-1);
}
else
if(one<0 && two <0)
{
return simpleMultiply(0-one,0-two);
}
else
if(one<0 && two >0)
{
return simpleMultiply(0-one,0-two);
}
else
if(one>0&&two<0)
{
return (0-one)+simpleMultiply(one,two+1);
}
return 0;
}

public class convert {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		System.out.println(multiply(10,300));
	}

	private static int multiply(int a, int b) {
		int result=0;
		if(a>b)
		{
			for(int i=1;i<=b;i++)
			{
				result= add(result,a);
			}
		}
		else
		{
			for(int i=1;i<=a;i++)
			{
				result= add(result,b);
			}
		}
		return result;
	}

	private static int add(int a, int b) {
		int carry=0;
		while(b!=0)
		{
			carry=a&b;
			a=a^b;
			b=carry<<1;
		}
		return a;
	}
}

def simpleMul(a,b)
  i = a
  (b-1).times do
    a += i
  end
  a
end
p simpleMul(4,8)

Very Simple Logic
Input A, B
--> Left Shift the value means multiply by 2
--> Add A the above result once if B is odd number

This solution works irrespective of signs.

public static int multiply(int a, int b){
		int product = 0;		
		for(int i=0; i<Math.abs(a); i++){		
			if(a<0){
				product = product - b;
			}else{
				product = product + b;
			}
		}	
		return product;

}

Laziest Solution:-

a=5, b=2
result= a / (1/b) = 5 / (1/ 2) = 10

Note: not using mulitiply operator

def mult(x, y)
  total = 0
  y.times do 
    total += x
  end
  total
end