Java番石榴 |带示例的 LongMath 类的 gcd(long a, long b)

Guava 的 LongMath 类的方法gcd(long a, long b)返回两个参数 a 和 b 的最大公约数。

句法：

public static long gcd(long a, long b)

参数：该方法接受两个参数a和b，分别是要查找其GCD的long类型。

返回类型：此方法返回将传递给函数的两个参数相除的最大正长值。

例外：如果 a 为负数或 b 为负数，方法 gcd(long a, long b) 将抛出IllegalArgumentException 。

注意：如果 a 和 b 都为零，则该方法返回零。

示例 1：

// Java code to show implementation of
// gcd(long a, long b) method of Guava's
// LongMath class
  
import java.math.RoundingMode;
import com.google.common.math.LongMath;
  
class GFG {
  
    // Driver code
    public static void main(String args[])
    {
        long a1 = 14;
        long b1 = 70;
  
        // Using gcd(long a, long b) method
        // of Guava's LongMath class
        long ans1 = LongMath.gcd(a1, b1);
  
        System.out.println("GCD of " + a1
                           + " and " + b1
                           + " is " + ans1);
  
        long a2 = 23;
        long b2 = 15;
  
        // Using gcd(long a, long b) method
        // of Guava's LongMath class
        long ans2 = LongMath.gcd(a2, b2);
  
        System.out.println("GCD of " + a2
                           + " and " + b2
                           + " is " + ans2);
    }
}

输出：

GCD of 14 and 70 is 14
GCD of 23 and 15 is 1

示例 2：

// Java code to show implementation of
// gcd(long a, long b) method of Guava's
// LongMath class
  
import java.math.RoundingMode;
import com.google.common.math.LongMath;
  
class GFG {
  
    // Driver code
    public static void main(String args[])
    {
        long a = -5;
        long b = 15;
  
        try {
            // Using gcd(long a, long b) method
            // of Guava's LongMath class
            // This should throw "IllegalArgumentException"
            // as a < 0
            long ans = LongMath.gcd(a, b);
        }
        catch (Exception e) {
            System.out.println(e);
        }
    }
}

输出：

java.lang.IllegalArgumentException: a (-5) must be >= 0

参考： https://google.github.io/guava/releases/20.0/api/docs/com/google/common/math/LongMath.html#gcd-long-long-