Question

) Design and test a function int mod2Division_4(int* np, int* mp) to display the remainder of n/m in modulo 2 arithmetic, where dividend n is the 4-bit binary number saved in an 4-element array pointed by np, and divisor m is the 4-bit binary number saved in an 4- element array pointed by mp. Save and display the remainder as binary digits in an array int remainder[4]. The function returns the number of digits in remainder starting from 1. For example, when n=11002 and m=10112, the remainder of n/m is n%m =1100 - 1011 = 0111, the function returns 3, and the program outputs 0 1 1 1. Test the function at least three times with different binary numbers

Answer 1

#include<bits/stdc++.h>
using namespace std;
int mod2division_4(vector<int> &np,vector<int> &mp)
{
   vector<int> remainder;
   for(int i=0;i<4;i++)
   {
       remainder.push_back(np[i]^mp[i]); //finding remainder by doing XOR operation
   }
  
  
   int count=0;
   for(int i=0;i<remainder.size();i++)
   {
       if(remainder[i]==1) //counting the set bits
       count++;
   }
  

   cout<<"Remainder is "<<" ";
   for(int i=0;i<4;i++)
   {
       cout<<remainder[i]<<" ";
   }
cout<<endl;
  
   return count;
  
}
int main()
{
   // First Input
   vector<int> np;
   for(int i=0;i<4;i++)
{
   int x;
   cin>>x;
   np.push_back(x);
   }
  
   //Second Input
   vector<int> mp;
   for(int i=0;i<4;i++)
{
   int x;
   cin>>x;
   mp.push_back(x);
   }
  
   int ans=mod2division_4(np,mp);
   cout<<"Number of digits in remainder starting from 1 are -> "<<ans<<endl;
  
   return 0;
}

The output of the above program with different Test Cases are as follows-