CareerCup

#include<stdio.h>
int main(void){

int a ,b,s=0,m,i=1;
printf(" \n Enter number 1 ");
scanf("%d",&a);
printf("Enter number 2 ");
scanf("%d",&b);
while(b>10)
{
m=b%10;
m=m*a*i; // 12x11 => 12 + 120
s=s+m;
b=b/10;
i=i*10;
scanf("%d",&s);
}
scanf("%d",&s);
return 0;
}

and

//multiplication as repeated addition
#include <stdio.h>
#include <stdlib.h>

int main()
{
int multiplicant,multiplier,i,result=0;
scanf("%d",&multiplicant);
scanf("%d",&multiplier);
for(i=0;i<multiplicant;i++)
{
result+=multiplier;

}
printf("result =%d",result);
return 0;
}

and

//multiplication as repeated addition
#include <stdio.h>
#include <stdlib.h>

int main()
{
int multiplicant,multiplier,i,result=0;
scanf("%d",&multiplicant);
scanf("%d",&multiplier);
for(i=0;i<multiplicant;i++)
{
result+=multiplier;

}
printf("result =%d",result);
return 0;
}

//multiplication as repeated addition
#include <stdio.h>
#include <stdlib.h>

int main()
{
    int multiplicant,multiplier,i,result=0;
    scanf("%d",&multiplicant);
    scanf("%d",&multiplier);
    for(i=0;i<multiplicant;i++)
    {
        result+=multiplier;

    }
    printf("result =%d",result);
    return 0;
}

are numbers integers or could they be floats or doubles?

for integers the solution is simple and obvious,
what if the input numbers are doubles or floats?

With less complexity
#include <stdio.h>
#include <stdlib.h>

int main()
{
int multiplicant,multiplier,i,result=0;
scanf("%d",&multiplicant);
scanf("%d",&multiplier);
for(i=0;i<multiplicant/2;i++)
{
result+=multiplier;

}
result+=result;
if(multiplicant%2 !=0){
result+=multiplier;
}
printf("result =%d",result);
return 0;
}

its not working

For integers

#include <stdio.h>
#include <stdlib.h>

int multiply(int multiplier, int multiplicant);

int main()
{
    int multiplicant,multiplier,i,result=0;
  scanf("%d",&multiplicant);
  scanf("%d",&multiplier);
    printf(" result =%d", multiply(multiplier,multiplicant));
    return 0;
}

int multiply(int multiplier, int multiplicant){
    if(multiplicant == 0 || multiplicant ==1){
        return multiplier;
    }
    int result = multiply(multiplier, multiplicant/2);
    result +=result;
    if( multiplicant %2 ==0){
        return result;
    } else{
        return result+multiplier;
    }    
}

def mult(a, b):
	sum = 0
	while b != 0:
		if b & 1 != 0:
			sum = sum + a
		b >>= 1
		a <<= 1
	return sum
print(mult(44, 88))

I think the tricky part is that you need to iterate for the small number, also take care of the sign. C# code

public int Multiply(int a, int b)
{
	int sign = a > 0 ^ b > 0 ? -1 : 1;

	a = Math.Abs(a);
	b = Math.Abs(b);
	int n = Math.Min(a, b);
	int number = Math.Max(a, b);
	int total = 0;

	for (int i = 0; i < n; i++)
		total += number;

	return sign * total;
}

void mulAsAddition(int n,int m)
{
	int iProduct =0;
	for(int i=0;i<m;i++)
		iProduct=iProduct+n;
	cout<<endl<< "product of" << n << "and" << m<<"is"<<iProduct;
}

#include <stdio.h>
int main()
{
long m1,m2,res=0;
printf("enter the first number\n");
scanf("%d",&m1);
printf("enter the second number\n");
scanf("%d",&m2);int i;
for(i=0;i<m1;i++)
res=res+m2;
printf("%d",res);
return 0;
}