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: 1024
12 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-