Problem 1 (Text book, problem 4, programming projects, chapter 2, page 158) Weir

Question

Problem 1 (Text book, problem 4, programming projects, chapter 2, page 158) Weird multiplication For this assignment, you will be implementing the so-called "Russian Peasant" or "Ancient Egyptian method for multiplication. It looks a little odd, but just think of it as an algorithm, a recipe for doing multiplication in a way other than what you learned in grade school. The algorithm is as follows. If A and B are the 2 integers (only integers) to be multiplied, we repeatedly multiply A by 2 and divide B by 2, until B cannot be divided any further, that is, until its value becomes 0 (remember, this is integer division). During each step, whenever B is an odd number, we add the corresponding A value to the product we are generating. In the end, the sum of the A values that had corresponding odd B values is the product. Get it? Here is an example If the two integers to be multiplied are 34 and 19, the operations would be: Comment 34 68 136 272 544 19 Add A to the product, B is odd 9 Add A to the product, B is odd 4 Ignore this A value, B is even 2 Ignore this A value, B is evern 1 Add A to the product, B is odd Sum up all the A values that had odd B values and you get: 34+68+544- 646>Final product. (a) Part 1: Write a program to find the product of two integers (b) Part 2: Modify your program so that it repeatedly asks whether you want to find another product.

Explanation / Answer

#include<iostream>

using namespace std;

int main(){

cout<<"Hello to Weird multiplication calculator"; // Commented section is for part 2 of the question.

char ch= 'y';

int sum=0;

int n1;

int n2;

//do{

cout<<endl<<"Enter number 1: ";

cin>>n1;

cout<<endl<<"Enter number 2: ";

cin>>n2;

while(n2){

if(n2%2)

{

sum+=n1;

}

n1*=2;

n2/=2;

}

cout<<endl<<"Product = "<<sum;

cout<<endl<<endl<<"Do you want to enter new numbers ?";

cout<<"Do you want to enter more ?(y/n)";

cin>>ch;

//}

//while(ch=='y'||ch=='Y');

}

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