# function to return gcd of a and b
# Taking the matrix as globally
dp = [[-1 for i in range(1001)] for j in range(1001)]
def gcd(a,b):
# Everything divides 0
if (a == 0):
return b
if (b == 0):
return a
# base case
if (a == b):
return a
if(dp[a][b] != -1):
return dp[a][b]
# a is greater
if (a > b):
dp[a][b] = gcd(a-b, b)
else:
dp[a][b] = gcd(a, b-a)
return dp[a][b]
# Driver program to test above function
a = 98
b = 56
if(gcd(a, b)):
print('GCD of', a, 'and', b, 'is', gcd(a, b))
else:
print('not found')
# This code is contributed by Samim Hossain Mondal.
// Java program to find GCD of two numbers
import java.util.*;
public class GFG
{
static int [][]dp = new int[1001][1001];
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// if a value is already
// present in dp
if(dp[a][b] != -1)
return dp[a][b];
// a is greater
if (a > b)
dp[a][b] = gcd(a-b, b);
// b is greater
else
dp[a][b] = gcd(a, b-a);
// return dp
return dp[a][b];
}
// Driver method
public static void main(String[] args)
{
for(int i = 0; i < 1001; i++) {
for(int j = 0; j < 1001; j++) {
dp[i][j] = -1;
}
}
int a = 98, b = 56;
System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));
}
}
// This code is contributed by Samim Hossain Mondal.
import math
# Using the built-in gcd method
gcd_built_in = math.gcd(10, 5)
print("GCD:", gcd_built_in) # GCD: 5
# Creating a new gcd method
# implementing Euclid's algorithm
def gcd_euclid(num1, num2):
if(num1 == 0):
return num2
if(num2 == 0):
return num1
while(num1 != num2):
if(num1 > num2):
num1 = num1-num2
else:
num2 = num2 - num1
return num1
gcd_euclid = gcd_euclid(3, 7)
print("GCD:", gcd_euclid) # GCD: 1
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#Codechef Problem Solution Video: https://cutt.ly/codeChefContestProblemSolution
import math
num = [2,3,4,5,6,7]
res = num[0]
for i in range(1, len(num), 1):
res = math.gcd(res, num[i])
print(res)