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

• Difficulty Level : Medium
• Last Updated : 13 Jul, 2020

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.

Examples:

Input: str = “54326789013892014531903492543267890138920145319034925432678901389201”
Output: 54326789013892014531903492543267890138920145319034925432678901389201

Input: str = “7890138920145319034925432678907890138920145319034925432678901903492543267890”
Output: 7890138920145319034925432678907890138920145319034925432678901903492543267890

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
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++

 `// C++ program for the above approach``#include ``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'``;``        ``a++;``    ``}`` ` `    ``// 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;``}`
Output:
```12345678098765431234567809876543
```

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

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