两个大数的LCM

给定两个大数’a’和’b’使得(10 ^ 20 <= a，b <= 10 ^ 300)。找到给定的两个大数字的最小公倍数。
例子：

Input: a = 234516789234023485693020129
        b = 176892058718950472893785940
 Output: 41484157651764614525905399263631111992263435437186260

 Input: a = 36594652830916364940473625749407
        b = 448507083624364748494746353648484939
 Output: 443593541011902763984944550799004089258248037004507648321189937329

解决方案：在给定的问题中，我们可以看到数量非常大，超出了所有可用原始数据类型的限制，因此我们必须使用Java的BigInteger Class的概念。因此，我们将给定的字符串转换为biginteger，然后使用Java.math.BigInteger.gcd(BigInteger val)方法计算大数的gcd，然后使用以下公式计算lcm：
LCM * HCF = x * y，其中x和y是两个数字

以下是上述想法的实现。

Java

// Java Program to find LCM of two large numbers
import java.math.*;
import java.lang.*;
import java.util.*;
  
public class GFG {
  
    // function to calculate LCM of two large numbers
    public static BigInteger lcm(String a, String b)
    {
        // convert string 'a' and 'b' into BigInteger
        BigInteger s = new BigInteger(a);
        BigInteger s1 = new BigInteger(b);
  
        // calculate multiplication of two bigintegers
        BigInteger mul = s.multiply(s1);
  
        // calculate gcd of two bigintegers
        BigInteger gcd = s.gcd(s1);
  
        // calculate lcm using formula: lcm * gcd = x * y
        BigInteger lcm = mul.divide(gcd);
        return lcm;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
  
        // Input 'a' and 'b' are in the form of strings because
        // they can not be handled by integer data type
        String a = "36594652830916364940473625749407";
        String b = "448507083624364748494746353648484939";
  
        System.out.print(lcm(a, b));
    }
}
// Code contributed by Saurav Jain

Python3

# Python3 program to find LCM of two 
# large numbers
import math
  
# Function to calculate LCM of two
# large numbers
def lcm (a, b):
  
    # Convert string 'a' and 'b' 
    # into Integer
    s = int(a)
    s1 = int(b)
  
    # Calculate multiplication of 
    # both integers
    mul = s * s1
  
    # Calculate gcd of two integers
    gcd = math.gcd(s, s1)
  
    # Calculate lcm using 
    # formula: lcm * gcd = x * y
    lcm = mul // gcd
  
    return lcm
  
# Driver Code
if __name__ == '__main__':
  
    # Input 'a' and 'b' are in the
    # form of strings
    a = "36594652830916364940473625749407"
    b = "448507083624364748494746353648484939"
  
    print(lcm(a, b))
  
# This code is contributed by himanshu77

输出：

443593541011902763984944550799004089258248037004507648321189937329