# GCD of two numbers when one of them can be very large

• Difficulty Level : Medium
• Last Updated : 31 Aug, 2021

Given two numbers ‘a’ and ‘b’ such that (0 <= a <= 10^12 and b <= b < 10^250). Find the GCD of two given numbers.
Examples :

```Input: a = 978
b = 89798763754892653453379597352537489494736
Output: 6

Input: a = 1221
b = 1234567891011121314151617181920212223242526272829
Output: 3```

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Solution : In the given problem, we can see that first number ‘a’ can be handled by long long int data type but second number ‘b’ can not be handled by any int data type. Here we read second number as a string and we will try to make it less than and equal to ‘a’ by taking it’s modulo with ‘a’.
Below is implementation of the above idea.

## C++

 `// C++ program to find GCD of two numbers such that``// the second number can be very large.``#include``using` `namespace` `std;``typedef` `long` `long` `int` `ll;` `// function to find gcd of two integer numbers``ll gcd(ll a, ll b)``{``    ``if` `(!a)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Here 'a' is integer and 'b' is string.``// The idea is to make the second number (represented``// as b) less than and equal to first number by``// calculating its mod with first integer number``// using basic mathematics``ll reduceB(ll a, ``char` `b[])``{``    ``// Initialize result``    ``ll mod = 0;` `    ``// calculating mod of b with a to make``    ``// b like 0 <= b < a``    ``for` `(``int` `i = 0; i < ``strlen``(b); i++)``        ``mod = (mod * 10 + b[i] - ``'0'``) % a;` `    ``return` `mod; ``// return modulo``}` `// This function returns GCD of 'a' and 'b'``// where b can be very large and is represented``// as a character array or string``ll gcdLarge(ll a, ``char` `b[])``{``    ``// Reduce 'b' (second number) after modulo with a``    ``ll num = reduceB(a, b);` `    ``// gcd of two numbers``    ``return` `gcd(a, num);``}` `// Driver program``int` `main()``{``    ``// first number which is integer``    ``ll a = 0;` `    ``// second number is represented as string because``    ``// it can not be handled by integer data type``    ``char` `b[] = ``"1234567891011121314151617181920212223242526272829"``;``    ``if` `(a == 0)``        ``cout << b << endl;``    ``else``        ``cout << gcdLarge(a, b) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to find``// GCD of two numbers``// such that the second``// number can be very large.` `class` `GFG``{` `    ``// This function computes``    ``// the gcd of 2 numbers``    ``private` `static` `int` `gcd(``int` `reduceNum, ``int` `b)``    ``{``        ``return` `b == ``0` `?``            ``reduceNum : gcd(b, reduceNum % b);``    ``}` `    ``// Here 'a' is integer and 'b'``    ``// is string. The idea is to make``    ``// the second number (represented``    ``// as b) less than and equal to``    ``// first number by calculating its``    ``// modulus with first integer``    ``// number using basic mathematics``    ``private` `static` `int` `reduceB(``int` `a, String b)``    ``{``        ``int` `result = ``0``;``        ``for` `(``int` `i = ``0``; i < b.length(); i++)``        ``{``            ``result = (result * ``10` `+``                      ``b.charAt(i) - ``'0'``) % a;``        ``}``        ``return` `result;``    ``}` `    ``private` `static` `int` `gcdLarge(``int` `a, String b)``    ``{``        ``// Reduce 'b' i.e the second``        ``// number after modulo with a``        ``int` `num = reduceB(a, b);``        ` `        ``// Now,use the euclid's algorithm``        ``// to find the gcd of the 2 numbers``        ``return` `gcd(num, a);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// First Number which``        ``// is the integer``        ``int` `a = ``1221``;``        ` `        ``// Second Number is represented``        ``// as a string because it cannot``        ``// be represented as an integer``        ``// data type``        ``String b = ``"19837658191095787329"``;``        ``if` `(a == ``0``)``            ``System.out.println(b);``        ``else``            ``System.out.println(gcdLarge(a, b));``    ``}` `// This code is contributed``// by Tanishq Saluja.``}`

## Python3

 `# Python3 program to find GCD of``# two numbers such that the second``# number can be very large.`  `# Function to find gcd``# of two integer numbers``def` `gcd(a, b) :``    ` `    ``if` `(a ``=``=` `0``) :``        ``return` `b``        ` `    ``return` `gcd(b ``%` `a, a)` `# Here 'a' is integer and 'b' is string.``# The idea is to make the second number``# (represented as b) less than and equal``# to first number by calculating its mod``# with first integer number using basic``# mathematics``def` `reduceB(a, b) :``    ` `    ``# Initialize result``    ``mod ``=` `0` `    ``# Calculating mod of b with a``    ``# to make b like 0 <= b < a``    ``for` `i ``in` `range``(``0``, ``len``(b)) :``        ` `        ``mod ``=` `(mod ``*` `10` `+` `ord``(b[i])) ``%` `a` `    ``return` `mod      ``# return modulo`  `# This function returns GCD of``# 'a' and 'b' where b can be``# very large and is represented``# as a character array or string``def` `gcdLarge(a, b) :``    ` `    ``# Reduce 'b' (second number)``    ``# after modulo with a``    ``num ``=` `reduceB(a, b)` `    ``# gcd of two numbers``    ``return` `gcd(a, num)`  `# Driver program` `# First number which is integer``a ``=` `1221` `# Second number is represented``# as string because it can not``# be handled by integer data type``b ``=` `"1234567891011121314151617181920212223242526272829"``if` `a ``=``=` `0``:``    ``print``(b)``else``:``    ``print``(gcdLarge(a, b))`  `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find GCD of``// two numbers such that the``// second number can be very large.``using` `System;` `class` `GFG``{``// function to find gcd``// of two integer numbers``public` `long` `gcd(``long` `a, ``long` `b)``{``    ``if` `(a == 0)``    ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Here 'a' is integer and``// 'b' is string. The idea``// is to make the second``// number (represented as b)``// less than and equal to``// first number by calculating``// its mod with first integer``// number using basic mathematics``public` `long` `reduceB(``long` `a, ``string` `b)``{``    ``// Initialize result``    ``long` `mod = 0;` `    ``// calculating mod of``    ``// b with a to make``    ``// b like 0 <= b < a``    ``for` `(``int` `i = 0; i < b.Length; i++)``        ``mod = (mod * 10 +``              ``(b[i] - ``'0'``)) % a;` `    ``return` `mod;``}` `// This function returns GCD``// of 'a' and 'b' where b can``// be very large and is``// represented as a character``// array or string``public` `long` `gcdLarge(``long` `a, ``string` `b)``{``    ``// Reduce 'b' (second number)``    ``// after modulo with a``    ``long` `num = reduceB(a, b);` `    ``// gcd of two numbers``    ``return` `gcd(a, num);``}` `// Driver Code``static` `void` `Main()``{``    ``// first number``    ``// which is integer``    ``long` `a = 1221;` `    ``// second number is represented``    ``// as string because it can not``    ``// be handled by integer data type``    ``string` `b = ``"1234567891011121314151617181920212223242526272829"``;``    ``GFG p = ``new` `GFG();``    ``if` `(a == 0)``        ``Console.WriteLine(b);``    ``else``        ``Console.WriteLine(p.gcdLarge(a, b));` `}``}` `// This code is contributed by mits.`

## PHP

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## Javascript

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Output :

`3`