5 Divisiblity不使用％，和/操作者 [英] Divisiblity of 5 without using % and / operator

问题描述

如何检查一个数是否整除5，并无须使用％和/运营商。 我想一个最快的算法，对于这个问题。

how to check whether a number is divisible by 5 or not without using % and / operator. I want a quickest algorithm for this problem.

推荐答案

一个很好的出发点是要考虑如何划分可以完成与乘法和位的变化。 这个问题是一个地方看。

A good starting point is to look into how division can be accomplished with multiplication and bit-shifts. This question is one place to look.

在特定的，可以按照连接后在以下策略来打。首先，用乘法和位移除以5：

In particular, you can follow the attached post to hit upon the following strategy. First, "divide by 5" using multiplication and bit-shifts:

 int32_t div5(int32_t dividend) {
     int64_t invDivisor = 0x33333333;
     return 1 + (int32_t) ((invDivisor * dividend) >> 32);
 }

然后，把结果乘以5：

Then, take the result and multiply by 5:

int result = div5(dividend) * 5;

然后，结果==分红，当且仅派息是整除5。

Then, result == dividend if and only dividend is divisible by 5.

if(result == dividend) {
    // dividend is divisible by 5
}
else {
    // dividend is not divisible by 5
}

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