Java 数值计算

java.lang.Math's floor(),ceil() and round()的区别：
Just 贴 JDK：

引用

public static double floor(double a)
    Returns the largest (closest to positive infinity) double value that is less than or equal to the argument and is equal to a mathematical integer.
floor返回的是不大于(小于or等于)参数的最大整数的floating-point value表示。

public static double ceil(double a)
    Returns the smallest (closest to negative infinity) double value that is greater than or equal to the argument and is equal to a mathematical integer.
ceil返回的是不小于(大于or等于)参数的最小整数的floating-point value表示。

public static long round(double a)
    Returns the closest long to the argument. The result is rounded to an integer by adding 1/2, taking the floor of the result, and casting the result to type long. In other words, the result is equal to the value of the expression:
    (long)Math.floor(a + 0.5d)
在做四舍五入运算时，我们可以使用round;但当参数为 -xxx.5 时，四舍五入后得到的结果不是我们期待的，解决办法是先对参数取绝对值，然后再用round方法

Sources：
http://blog.csdn.net/foart/archive/2009/06/24/4295645.aspx


使用BigDecimal做精确的数值计算：
Java精确计算 - http://blog.csdn.net/xiaosu_521/article/details/2065854



关于round方法结合面试题再多说说：
Math.round(11.5) = 12
Math.round(-11.5) = -11
Math.round(-11.3) = -11
Math.round(-11.8) = -12
析：

引用

Math.round(11.5)
     = (long)Math.floor(11.5 + 0.5)
     = (long)Math.floor(12.0)
     = (long)12.0
     = 12
    
Math.round(-11.5)
     = (long)Math.floor(-11.5 + 0.5)
     = (long)Math.floor(-11.0)
     = (long)-11.0
     = -11    
    
Math.round(-11.3)
     = (long)Math.floor(-11.3 + 0.5)
     = (long)Math.floor(-10.8)
     = (long)-11.0
     = -11
    
Math.round(-11.8)
     = (long)Math.floor(-11.8 + 0.5)
     = (long)Math.floor(-11.3)
     = (long)-12.0
     = -12