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waiting 4……what? | Just another WordPress.com site | iafita.wordpress.com Reviews
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Just another WordPress.com site
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1 skip to content
2 早前文章
3 hello world
4 posted on
5 afitalive
6 uncategorized
7 一条评论
8 一、问题描述
9 求一个n位整数x的二进制表示中1的个数,越快越好
10 据说这是道很常见的面试题
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skip to content,早前文章,hello world,posted on,afitalive,uncategorized,一条评论,一、问题描述,求一个n位整数x的二进制表示中1的个数,越快越好,据说这是道很常见的面试题,原题是说如何在常量时间内算出32位整数的二进制表示中1的个数,但如果你这么告诉面试官,十有八九会被骂白痴,因为大家一般都会认为32次实际上是o n 的做法,所以这么问可能会更好一些,求一个n位整数x的二进制表示中1的个数,越快越好,另外在信息论、编码学等也有很多应用,算术运算
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waiting 4……what? | Just another WordPress.com site | iafita.wordpress.com Reviews

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九月 | 2008 | waiting 4……what?

https://iafita.wordpress.com/2008/09

Waiting 4……what? Just another WordPress.com site. Ubuntu Is *NOT* Hard Drive Killer. 九月 13, 2008. Ubuntu 中的Load/Unload Cycle Count问题及解决方案 … 继续阅读 →. Popcount Problem: 求二进制数中1的个数 (上). Popcount Problem: 求二进制数中1的个数 (下). Ubuntu Is *NOT* Hard Drive Killer. Popcount Problem: 求二进制数中1的个数…. 上的 Pop Count Problem (二…. Waiting 4……what? 通过访问 WordPress.com 创建免费网站或博客.

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Popcount Problem: 求二进制数中1的个数 (下) | waiting 4……what?

https://iafita.wordpress.com/2009/01/30/popcount-problem-求二进制数中1的个数-下

Waiting 4……what? Just another WordPress.com site. Popcount Problem: 求二进制数中1的个数 (上) →. Popcount Problem: 求二进制数中1的个数 (下). 一月 30, 2009. 总共需要3次shift, 3次and, 2次add, 1次mul, 1次mod 共. 2=4*a 2*b c,x 1&m1= 2*a c,x 2&m2=a,于是 x – x 1&m1 – x 2&m2 = a b c。 二是在第二步做完之后,这时是每6位自成一组,以12位整数为例,x=(ab) 64 = a*64 b。 很明显, a*64 b. A b mod 63。 6) 再次优化shift and add方法,用乘法操作代替加法操作. 总共需要 4次shift, 4次and, 2次加法,1次减法,1次乘法共 12. 0a 0b 0c 0d. 01 01 01 01. 8212;———-. 8212;———–. 2] Everything上的讨论: http:/ everything2.com/index.pl? You are comme...

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二月 | 2011 | waiting 4……what?

https://iafita.wordpress.com/2011/02

Waiting 4……what? Just another WordPress.com site. 二月 13, 2011. Welcome to WordPress.com. This is your f … 继续阅读 →. Popcount Problem: 求二进制数中1的个数 (上). Popcount Problem: 求二进制数中1的个数 (下). Ubuntu Is *NOT* Hard Drive Killer. Popcount Problem: 求二进制数中1的个数…. 上的 Pop Count Problem (二…. Waiting 4……what?

4

十一月 | 2007 | waiting 4……what?

https://iafita.wordpress.com/2007/11

Waiting 4……what? Just another WordPress.com site. 十一月 28, 2007. 结果退休后,因为年纪大,身体差,行动不方便, … 继续阅读 →. 十一月 22, 2007. 女人入洞房那天,早早收起了自己的鞋,等男人脱鞋上炕,女人却双脚踩在男人的鞋上 … 继续阅读 →. 十一月 12, 2007. 有吧,结果也分好的坏的 … 继续阅读 →. 十一月 3, 2007. 与同学相处和适应 … 继续阅读 →. Popcount Problem: 求二进制数中1的个数 (上). Popcount Problem: 求二进制数中1的个数 (下). Ubuntu Is *NOT* Hard Drive Killer. Popcount Problem: 求二进制数中1的个数…. 上的 Pop Count Problem (二…. Waiting 4……what? 通过访问 WordPress.com 创建免费网站或博客.

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Popcount Problem: 求二进制数中1的个数 (上) | waiting 4……what?

https://iafita.wordpress.com/2009/01/30/popcount-problem-求二进制数中1的个数-上

Waiting 4……what? Just another WordPress.com site. Popcount Problem: 求二进制数中1的个数 (下). Popcount Problem: 求二进制数中1的个数 (上). 一月 30, 2009. 实际上这个问题叫做Hamming weight[1],或者叫做population count以及pop count。 对于这个问题的解法,Google一下popcount 或者count bits 1会有一大把。 Pop count lnaive (. 同理,x (x 1) 可以消除最右端的一个0(转化为1)。 Use this instead if most bits in x are 1 instead of 0. 3) shift and add 时间复杂度O(logN). 以4位整数abcd为例,最终结果是a b c d,循环的话需要4步加法。 1′ 每一位一组[a][b][c][d],相邻两组相加得到[a b][c d],结合成两位一组。 0] [b] [0] [d]. 0] [a] [0] [c]. E f] [g h].

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waiting 4……what? | Just another WordPress.com site

Waiting 4……what? Just another WordPress.com site. 二月 13, 2011. Welcome to WordPress.com. This is your first post. Edit or delete it and start blogging! Popcount Problem: 求二进制数中1的个数 (上). 一月 30, 2009. 实际上这个问题叫做Hamming weight[1],或者叫做population count以及pop count。 对于这个问题的解法,Google一下popcount 或者count bits 1会有一大把。 Pop count lnaive (. 同理,x (x 1) 可以消除最右端的一个0(转化为1)。 Use this instead if most bits in x are 1 instead of 0. 3) shift and add 时间复杂度O(logN). 以4位整数abcd为例,最终结果是a b c d,循环的话需要4步加法。 0] [b] [0] [d]. 0] [a] [0] [c].

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