Programming Exercise 11 in Chapter 8 explains how to add large integers using ar
ID: 3830396 • Letter: P
Question
Programming Exercise 11 in Chapter 8 explains how to add large integers using arrays. However, in that exercise, the program could add only integers of, at most, 20 digits. This chapter explains how to work with dynamic integers. Design a class named large integers such that an object of this class can store an integer of any number of digits. Add operations to add. subtract, multiply, and compare integers stored in two objects. Also add constructors to properly initialize objects and functions to set, retrieve, and print the values of objects. REFERENCE: (Adding Large Integers) In C + +, the largest int value is 2147483647. So, an integer larger than this cannot be stored and processed as an integer. Similarly, if the sum or product of two positive integers is greater than 2147483647, the result will be incorrect. One way to store and manipulate large integers is to store each individual digit of the number in an array. Write a program that inputs two positive integers of, at most, 20 digits and outputs the sum of the numbers. If the sum of the numbers has more than 20 digits, output the sum with an appropriate message. Your program must, at least, contain a function to read and store a number into an array and another function to output the sum of the numbers.Explanation / Answer
#include <iostream>
#include <cctype>
#include <cstring>
using namespace std;
#define DIGITS 20
class largeintegers
{
public:
largeintegers();
void Input();
void Output();
largeintegers operator+( largeintegers );
largeintegers operator-( largeintegers );
largeintegers operator*( largeintegers );
int operator==( largeintegers);
private:
int integer[ DIGITS ];
int len;
};
void largeintegers::Output()
{
int i;
for (i=len-1;i >= 0; i-- )
cout<<integer[i];
}
void largeintegers::Input()
{
string in;
int i,j,k;
cout << "Enter a number("<<DIGITS<<" digits max):";
cin >> in;
len = in.length();
k=0;
for(j=len-1;j>=0;j--){
integer[j]=in[k++]-48;
}
}
largeintegers::largeintegers( )
{
for ( int i = 0; i <DIGITS; i++ )
integer[ i ] = 0;
len=DIGITS-1;
}
int largeintegers::operator==( largeintegers op2 )
{int i;
if(len<op2.len)
return -1;
if(op2.len<len)
return 1;
for(i=len-1;i>=0;i--)
if(integer[i]<op2.integer[i])
return -1;
else if(op2.integer[i]<integer[i])
return 1;
return 0;
}
largeintegers largeintegers::operator+( largeintegers op2 )
{largeintegers temp;
int carry = 0;
int c,i;
if(len>op2.len)
c=len;
else
c=op2.len;
for ( i=0; i<c; i++ )
{temp.integer[ i ] =integer[ i ] + op2.integer[ i ] + carry;
if ( temp.integer[ i ] > 9 )
{temp.integer[ i ] %= 10;
carry = 1;
}
else
carry = 0;
}
if(carry==1)
{temp.len=c+1;
if(temp.len>=DIGITS)
cout<<"***OVERFLOW***** ";
else
temp.integer[i]=carry;
}
else
temp.len=c;
return temp;
}
largeintegers largeintegers::operator-( largeintegers op2 )
{largeintegers temp;
int c;
if(len>op2.len)
c=len;
else
c=op2.len;
int borrow = 0;
for( int i = c;i >= 0;i--)
if(borrow==0)
{if(integer[i]>=op2.integer[i])
temp.integer[i]=integer[i]-op2.integer[i];
else
{borrow=1;
temp.integer[i]=integer[i]+10-op2.integer[i];
}
}
else
{borrow=0;
if(integer[i]-1>=op2.integer[i])
temp.integer[i]=integer[i]-1-op2.integer[i];
else
{borrow=1;
temp.integer[i]=integer[i]-1+10-op2.integer[i];
}
}
temp.len=c;
return temp;
}
largeintegers largeintegers::operator*( largeintegers op2 )
{ largeintegers temp;
int i,j,k,tmp,m=0;
for (int i=0; i<op2.len; i++)
{ k=i;
for (j=0; j< len; j++)
{tmp = integer[ j ] * op2.integer[i];
temp.integer[k] =temp.integer[k]+tmp;
temp.integer[k+1] =temp.integer[k+1]+ temp.integer[k]/10;
temp.integer[k] %=10;
k++;
if(k>m)
m=k;
}
}
temp.len=m;
if(temp.len>DIGITS)
cout<<"***More than 20 digit! ERROR !***** ";
return temp;
}
using namespace std;
int main()
{int c;
largeintegers n1,n2,result;
n1.Input();
n2.Input();
n1.Output();
cout <<" + ";
n2.Output();
result=n1+n2;
cout<< " = " ;
result.Output();
cout << " ";
n1.Output();
cout <<" - ";
n2.Output();
result=n1-n2;
cout<< " = " ;
result.Output();
cout << " ";
n1.Output();
cout <<" * ";
n2.Output();
result=n1*n2;
cout<< " = " ;
result.Output();
cout << " ";
c=n1==n2;
n1.Output();
switch (c)
{case -1: cout<<" is less than ";
break;
case 0: cout<<" is equal to ";
break;
case 1: cout<<" is greater than ";
break;
}
n2.Output();
}
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