Guava – LongMath.pow方法与实例
Guava的LongMath类的方法pow(long b, int k)返回b的第k次方。即使结果溢出,它也将等于BigInteger.valueOf(b).pow(k).longValue()。这个实现在O(log k)时间内运行。
语法:
public static long pow(long b, int k)
参数: 该方法接受两个参数,b和k。参数b被称为base,它被提高到k次方.
返回值: 这个方法返回b的k-th次方。
异常: 如果k是负数,这个方法会抛出IllegalArgumentException。
示例1:
// Java code to show implementation of// pow(long b, int k) 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 b1 = 4; int k1 = 5; long ans1 = LongMath.pow(b1, k1); System.out.println(b1 + " to the " + k1 + "th power is: " + ans1); long b2 = 12; int k2 = 3; long ans2 = LongMath.pow(b2, k2); System.out.println(b2 + " to the " + k2 + "rd power is: " + ans2); }}
输出:
4 to the 5th power is: 102412 to the 3rd power is: 1728
示例2:
// Java code to show implementation of// pow(long b, int k) method of Guava's// LongMath class import java.math.RoundingMode;import com.google.common.math.LongMath; class GFG { static long findPow(long b, int k) { try { // Using pow(long b, int k) // method of Guava's LongMath class // This should throw "IllegalArgumentException" // as k < 0 long ans = LongMath.pow(b, k); // Return the answer return ans; } catch (Exception e) { System.out.println(e); return -1; } } // Driver code public static void main(String args[]) { long b = 4; int k = -5; try { // Using pow(long b, int k) // method of Guava's LongMath class // This should throw "IllegalArgumentException" // as k < 0 LongMath.pow(b, k); } catch (Exception e) { System.out.println(e); } }}
输出:
java.lang.IllegalArgumentException: exponent (-5) must be >= 0
参考: https://google.github.io/guava/releases/20.0/api/docs/com/google/common/math/LongMath.html#pow-long-int-
