计算在给定范围内偶数量的最简单方法 [英] Simplest way to calculate amount of even numbers in given range

问题描述

什么是计算在一个范围内的无符号整数的偶数量最简单的方法？

一个例子：如果范围是[0 ... 4]那么答案是3（0,2,4）

我有时间难以想到的任何简单的方法。唯一的解决方案，我想出了几个参与if语句的。有code的简单的线条，可以做到这一点没有if语句或三元操作？

解决方案

 诠释甚至=（0 ==开始％2）？ （结束 - 开始）/ 2 + 1：（结束 - 开始+ 1）/ 2;
 

哪些可转换成：

  INT即使=（结束 - 开始+（开头％2））/ 2 +（1  - （开头％2））;
 

编辑：可以进一步简化为：

  INT即使=（结束 - 开始+ 2  - （开头％2））/ 2;
 

EDIT2：由于对在我看来，用C整数除法有点不正确定义（整数除法向下截断为正数，向上为负数）时开始这个公式将无法工作是一个负奇数。

EDIT3：用户iPhone初学者正确地指出，如果开始％2 替换开始＆安培; 1 这将正常工作的所有范围。

What is the simplest way to calculate the amount of even numbers in a range of unsigned integers?

An example: if range is [0...4] then the answer is 3 (0,2,4)

I'm having hard time to think of any simple way. The only solution I came up involved couple of if-statements. Is there a simple line of code that can do this without if-statements or ternary operators?

解决方案

int even = (0 == begin % 2) ? (end - begin) / 2 + 1 : (end - begin + 1) / 2;

Which can be converted into:

int even = (end - begin + (begin % 2)) / 2 + (1 - (begin % 2));

EDIT: This can further simplified into:

int even = (end - begin + 2 - (begin % 2)) / 2;

EDIT2: Because of the in my opinion somewhat incorrect definition of integer division in C (integer division truncates downwards for positive numbers and upwards for negative numbers) this formula won't work when begin is a negative odd number.

EDIT3: User 'iPhone beginner' correctly observes that if begin % 2 is replaced with begin & 1 this will work correctly for all ranges.

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