Question

C++ pleasr

Extra Credit: Large Integer (25 points) 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. (Hint: Read numbers as strings and store the digits of the number in the reverse order.) Sample three runs Outputs for Extra Credit 01 a poStive tege 3456789 nter a 234567898765 stive integer of at most 20 digits: The sum of the numbers is: 34571355554 ress any key to continue ger of at most 20 1g1ts 12345678912345678912 Enter a postive integer of at most 20 digits: The sum of the numbers is: The sum of the numbers overflows. It has 21 digits 1112345678912345678911 ress any key to continue 12345678912345678912 This number has more than 20 digits. Program terminates e 10 points-a function to read the number 5 points will be deducted if you handle the overflow case 2.3 points will be deducted if you do not include at least one paragraph for analyzing the 15 points- a function to sum the two numbers .5 points will be deducted if you do not handle the more than 20 digits case problem and one paragraph for designing a solution .2.5 points will be deducted if you do not include comments within your code.

Answer 1

#include <iostream>
#include <sstream>
using namespace std;

string trimString(string str) {
std::stringstream trimmer;
trimmer << str;
trimmer >> str;
return str;
}

void readNum(int N[], int &length) {
string num;

do {
cout << "Enter the positive integer of at most 20 digits:";
cin >> num;

num = trimString(num);

if(num.size() > 20) {
cout << "Number should be of maximum 20 digits only." << endl;
}
} while(num.size() > 20);

length = num.size();

int count = 1;
for(int i = num.size() - 1; i >= 0; i--) {
N[20 - count++] = num.at(i) - '0';
}
}

void printNum(int N[], int arrayLength, int length) {
for(int i=arrayLength-length; i < arrayLength; i++) {
cout << N[i];
}
cout << endl;
}

void sumNum(int N1[], int l1, int N2[], int l2) {
int result[21] = {0};
int length = 0;

int max = (l1 > l2 ? l1 : l2);
int carry = 0;

for(int i=20-1; i>=(20 - max); i--) {
result[i + 1] = N1[i] + N2[i] + carry;

if(result[i + 1] >= 10) {
carry = 1;
result[i + 1] -= 10;
} else {
carry = 0;
}
length++;
}

cout << "The sum of the numbers is:" << endl;
if(carry != 0) {
result[21 - 1 - length] = 1;
length++;
}

if(result[0] != 0) {
cout << "the sum of the numbers overflows. It has 21 digits." << endl;
}
printNum(result, 21, length);
}

int main() {
int N1[20] = {0}, N2[20] = {0};
int l1, l2;
readNum(N1, l1);
readNum(N2, l2);

cout << endl;
cout << "First Number: ";
printNum(N1, 20, l1);
cout << "Second Number: ";
printNum(N2, 20, l2);

cout << endl;
sumNum(N1, l1, N2, l2);
}

saved share gcc version 4.6.3 .cpp Enter the positive integer of at most 20 digits: 98271387213652132153 Enter the positive integer of at most 20 digits: 98271387213652132153 39 40 41 void sumNum(int N1], int 11, int N2[], int 12) First Number: 98271387213652132153 Second Number: 9827138721 36521 32153 12 43 int result[21]-1o: int length 6; The sum of the numbers is: the sum of the numbers overflows. it has 21 digits 196542774427384264306 45 46 47 48- for(int i-20 1; i>-(20 max); i 49 50 51 52 53 54 int max-(11> 12 ? 11 :12); int carry0; result [ỉ l 11 N1[i] 1 N2[i] i carry; if(result[i + 1] >= 10) { carry 1; result[i 1]-10; else f carry ; 56 57 58 59 60 61- length++; cout << "The sum of the numbers is:" くくendl; if(carry !-0) 62 63 64 65 66- f(result[]-) result [21 1 length] 1; length++; cout << "the sum of the numbers overflows. 1t has 21 digits." << endl; 68 69 70 71 printNum(result, 21, length); 72- int main) t 73 74 75 int N1[20]eN2[20]-(o); int 11, 12; readNum(N1, 11)