C语言中长久的数学运算 [英] Math operations with long long in c

问题描述

给出 long long x ， long long y 和 int z ，如何截断(( x + y + z )/3)是在C语言中使用完全定义的运算来计算的，但不包括 unsigned long long 和更宽泛的算术运算吗?

Given long long x, long long y, and int z, how can trunc((x+y+z)/3) be calculated using fully defined operations in C but excluding unsigned long long and wider arithmetic?

(trunc( x )向零舍入":如果 x 是整数，则为 x ，否则为下一个整数趋近于零.)

(trunc(x) is "rounding toward zero": It is x if x is an integer and otherwise is the next integer toward zero.)

推荐答案

您可以计算两个整数 x 和 y 的平均值，而不会出现溢出风险，如下所示(伪代码):

You can calculate the average of two integers x and y without the risk of overflow as follows (pseudocode):

int average = (x and y have the same sign)
    ? x + (y - x) / 2
    : (x + y) / 2;

由于整数截断，这是不对称的.如果您想始终舍入为0，则需要做一些更复杂的事情:

This is not symmetric due to integer truncation. If you want to always round towards 0, you need to do something slightly more complicated:

int average = (x and y have the same sign)
    ? (x and y are both even or both odd)
          ? x + (y - x) / 2
          : x + (y - x - sgn(x)) / 2
    : (x + y) / 2;

其中 sgn 是符号函数(-1、0或+1，具体取决于参数是负数，零还是正数). sgn(x)的实现很多，但是一个不错的实现是:(x> 0)-(x< 0).

where sgn is the signum function (-1, 0, or +1 depending on whether the argument is negative, zero, or positive). There are lots of implementations of sgn(x), but one nice one is: (x > 0) - (x < 0).

如果您想始终四舍五入...则留作练习.:)

If you want to always round down...that's left as an exercise. :)

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