/******************************************* Author: Nautash Ahmad Created: 9th October, 2018 -------------------------------------------- Description: Create a program which will take two very long positive integers and outputs their sum. Both integers should have the same length (up to 100 digits). The challenge is to design and implement a custom data structure which will handle long integers properly. For example: Input: 54123512...(50 digits)...321 27198005...(37 digits)...483 Output: The sum of Num1 and Num2; -------------------------------------------- *******************************************/ #include #include #include using namespace std; int main() { string num1, num2; cin >> num1; cout << "First number: " << num1 << endl; cin >> num2; cout << "Second number: " << num2 << endl << endl; int length = 0; if(num1.size() >= num2.size()) length = num1.size() + 1; else length = num2.size() + 1; char *numArr1 = new char[length]; char *numArr2 = new char[length]; strcpy(numArr1, num1.c_str()); strcpy(numArr2, num2.c_str()); strrev(numArr1); strrev(numArr2); char *sumArr = new char[length + 2]; int carry = 0, temp = 0; memset(sumArr, ' ', length + 4); for(int i = 0; i < (length - 1); i++) { if(numArr1[i] < 48 || numArr1[i] > 57) numArr1[i] = '0'; if(numArr2[i] < 48 || numArr2[i] > 57) numArr2[i] = '0'; int num1 = numArr1[i] - '0'; int num2 = numArr2[i] - '0'; int sum = num1 + num2 + carry; sumArr[i] = (sum % 10) + '0'; carry = sum / 10; temp = i; } if((carry + '0') >= 48 && (carry + '0') < 58) sumArr[temp+1] = carry + '0'; strrev(sumArr); cout << "Sum of both numbers is: " << sumArr << endl; delete []numArr1; delete []numArr2; delete []sumArr; return 0; }

Long Adder