cn.ponfee.commons.math

类 Maths

java.lang.Object

cn.ponfee.commons.math.Maths

public class Maths
extends Object

数学算术 取模：Modulo Operation

作者:

Ponfee

构造器概要

构造器

构造器和说明
Maths()

方法概要

限定符和类型	方法和说明
static int	abs(int a)
static long	abs(long a)
static long	bitsMask(int bits) Returns a long value of bit count mask calculate the bit counts mask long value a: (1 << bits) - 1 b: -1L ^ (-1L << bits) c: ~(-1L << bits) d: Long.MAX_VALUE >>> (63 - bits) bitsMask(0) -> 0 -> 0000000000000000000000000000000000000000000000000000000000000000 bitsMask(1) -> 1 -> 0000000000000000000000000000000000000000000000000000000000000001 bitsMask(2) -> 3 -> 0000000000000000000000000000000000000000000000000000000000000011 bitsMask(10) -> 1023 -> 0000000000000000000000000000000000000000000000000000001111111111 bitsMask(20) -> 1048575 -> 0000000000000000000000000000000000000000000011111111111111111111 bitsMask(63) -> 9223372036854775807 -> 0111111111111111111111111111111111111111111111111111111111111111 bitsMask(64) -> -1 -> 1111111111111111111111111111111111111111111111111111111111111111
static int	gcd(int[] array) Returns the greatest common divisor in array
static int	gcd(int a, int b) Returns the greatest common divisor
static double	log(double n, double base) 求以base为底n的对数 Math.log10(double) 求以10为底n的对数（lg） Math.log(double) 以e为底n的对数（自然对数，ln） Math.log1p(double) 以e为底n+1的对数
static double	log2(double n) 以2为底n的对数
static int	minus(int a, int b)
static long	minus(long a, long b)
static int	plus(int a, int b)
static long	plus(long a, long b)
static long	pow(long base, int exponent) Returns a long value for baseexponent.
static int	rotateLeft(int x, int n) rotate shift left，循环左移位操作：0<=n<=32
static double	sqrtBinary(double value) Returns square root of specified double value Use binary search method
static double	sqrtNewton(double value) Returns square root of specified double value Use newton iteration method: X(n+1)=[X(n)+p/Xn]/2

从类继承的方法 java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

构造器详细资料

Maths

public Maths()

方法详细资料

log2

public static double log2(double n)

以2为底n的对数

参数:

n - the value

返回:

a value of log(n)/log(2)

log

public static double log(double n,
                         double base)

求以base为底n的对数 Math.log10(double) 求以10为底n的对数（lg） Math.log(double) 以e为底n的对数（自然对数，ln） Math.log1p(double) 以e为底n+1的对数

参数:

n - a value

base - 底数

返回:

a double of logarithm

rotateLeft

public static int rotateLeft(int x,
                             int n)

rotate shift left，循环左移位操作：0<=n<=32

参数:

x - the value

n - shift bit len

返回:

a number of rotate left result

bitsMask

public static long bitsMask(int bits)

 Returns a long value of bit count mask
 calculate the bit counts mask long value
   a: (1 << bits) - 1
   b: -1L ^ (-1L << bits)
   c: ~(-1L << bits)
   d: Long.MAX_VALUE >>> (63 - bits)

  bitsMask(0)  -> 0                   -> 0000000000000000000000000000000000000000000000000000000000000000
  bitsMask(1)  -> 1                   -> 0000000000000000000000000000000000000000000000000000000000000001
  bitsMask(2)  -> 3                   -> 0000000000000000000000000000000000000000000000000000000000000011
  bitsMask(10) -> 1023                -> 0000000000000000000000000000000000000000000000000000001111111111
  bitsMask(20) -> 1048575             -> 0000000000000000000000000000000000000000000011111111111111111111
  bitsMask(63) -> 9223372036854775807 -> 0111111111111111111111111111111111111111111111111111111111111111
  bitsMask(64) -> -1                  -> 1111111111111111111111111111111111111111111111111111111111111111
 

参数:

bits - the bit count

返回:

a long value

pow

public static long pow(long base,
                       int exponent)

Returns a long value for baseexponent.

参数:

base - the base

exponent - the exponent

返回:

a long value for baseexponent.

abs

public static int abs(int a)

abs

public static long abs(long a)

sqrtBinary

public static double sqrtBinary(double value)

Returns square root of specified double value

Use binary search method

参数:

value - the value

返回:

square root

sqrtNewton

public static double sqrtNewton(double value)

Returns square root of specified double value

Use newton iteration method: X(n+1)=[X(n)+p/Xn]/2

参数:

value - the value

返回:

square root

plus

public static int plus(int a,
                       int b)

minus

public static int minus(int a,
                        int b)

plus

public static long plus(long a,
                        long b)

minus

public static long minus(long a,
                         long b)

gcd

public static int gcd(int a,
                      int b)

Returns the greatest common divisor

参数:

a - the first number

b - the second number

返回:

gcd

gcd

public static int gcd(int[] array)

Returns the greatest common divisor in array

参数:

array - the int array

返回:

gcd