• ADDING TWO NUMBERS UP TO 10000 DIGITS

  /*We know that by using int ( integer type ) variables we can store it a number up to 9 or 10 digits.Even long int type can store up to 18 or 19 digits.So we can not store or add two numbers greater than this digits.No problem.Here is a simple program that could add two numbers up to 10000 digits.*/

   #include <iostream>
  
  using namespace std;
  
  int main()
  {
      string  x,y;
  
      int a[10000],b[10000],C[10000],k = 0;
  
      int counter,counter2;
  
      cin >> x >> y;
  
      if( x.length() >= y.length() ){
  
      for( counter = 0; counter < x.length(); counter++ )
  
          a[counter] = x[counter] - 48;
  
      for( counter2 = 0; counter2 < y.length(); counter2++ )
  
          b[counter2] = y[counter2] - 48;
      }
  
      if( x.length() < y.length() ){
  
      for( counter = 0; counter < y.length(); counter++)
  
          a[counter] = y[counter] - 48;
  
      for( counter2 = 0; counter2 < x.length(); counter2++)
  
          b[counter2] = x[counter2] - 48;
      }
      int rem = 0;
  
      for( int i = counter-1, j = counter2-1; i >=0 || j>=0; i--, j--, k++ )
      {
          if( i>=0 && j>=0 )
          {
              int c = a[i]+b[j] + rem;
  
              if(c < 10)
              {
                      rem = 0;
                      C[k] = c;
              }
              else
              {
                      c = c-10;
                      rem = 1;
                     C[k] = c;
  
              }
  
  
          }
      else
      {
          int c = a[i]+rem;
            if(c<10)
              {
                      rem = 0;
                      C[k] = c;
              }
              else
              {
                      c = c-10;
                      rem = 1;
                     C[k] = c;
  
              }
  
  
      }
  
      }
      if(rem == 1)
          C[k++] = rem;
          for(int i = k-1; ;i--)
          {
              if(C[i] == 0){
                  k--;
              }
              else
              {
                  break;
          }
          }
      for(int i = k-1; i >= 0; i--)
  
          cout << C[i];
  
          cout << endl;
  
  return 0;
  }