How to store a very large number of more than 100 digits in C++

Given an integer N in form of string str consisting of more than 100 digits, the task is to store the value for performing an arithmetic operation and print the given integer.


Input: str = “54326789013892014531903492543267890138920145319034925432678901389201”
Output: 54326789013892014531903492543267890138920145319034925432678901389201

Input: str = “7890138920145319034925432678907890138920145319034925432678901903492543267890”
Output: 7890138920145319034925432678907890138920145319034925432678901903492543267890

No data type is present in C++ to store 10100. So, the idea is to use get the input as string (as string can be of any length) and then convert this string into an array of digits of the length same as the length of string. Storing the big integer into an integer array will help to perform some basic arithmetic on that number.

Below are the steps:

  1. Take the large number as input and store it in a string.
  2. Create an integer array arr[] of length same as the string size.
  3. Iterate over all characters (digits) of string str one by one and store that digits in the corresponsing index of the array arr

    arr[i] = str[i] – ‘0’;

    // Here ‘0’ represents the digit 0, and
    // str[i] – ‘0’ = ASCII(str[i]) – ASCII(‘0’) = ASCII(str[i] – 48

  4. Using the above step, we can store very very large number for doing any arithmetic operations.

Below is the implementation of the above approach:






// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to return dynamic allocated
// array consisting integers individually
int* GetBigInteger(string str)
    int x = str.size(), a = 0;
    // Create an array to store the big
    // integer into it.
    // Make the array size same as the
    // size of string str
    int* arr = new int[str.size()];
    // Loop to extract string elements
    // into the array one by one
    while (a != x) {
        // Subtracting '0' to convert
        // each character into digit
        // str[a] - '0'
        // = ASCII(str[a]) - ASCII('0')
        // = ASCII(str[a] - 48
        arr[a] = str[a] - '0';
    // Return the reference of the array
    return arr;
// Driver Code
int main()
    // Big Integer in form of string str
    string str = "12345678098765431234567809876543";
    // Function Call
    int* arr = GetBigInteger(str);
    // Print the digits in the arr[]
    for (int i = 0; i < str.size(); i++) {
        cout << arr[i];
    return 0;




Time Complexity: O(K), K is the number of digits in the number
Auxiliary Space: O(K), K is the number of digits in the number

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Recommended Posts:

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.