bits专题

[LeetCode] 338. Counting Bits

题：https://leetcode.com/problems/counting-bits/description/ 题目 Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary repres

[LeetCode] 190. Reverse Bits

题：https://leetcode.com/problems/reverse-bits/ 题目大意将32位的数，二进制翻转。解题思路解法和将int a =123，翻转的思路相同。 int b= 0;while(a>0){b = b*10 + a %10;a /=10;} 将新数整体左移一位，然后取每个数的第i位置，将该位放到新数的最低位。循环32次遍可以得到翻转。

leetcode338 Counting Bits Java

Description Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array. Example: Fo

win7 x64 PL/SQL连接Make sure you have the 32 bits Oracle Cient italled.问题解决

一、当前环境 oracle11g win7 x64 PLSQL Developer 9.0.0.1601 使用plsql登录出现提示 Initialization error Could not initialize "F:\app\Administrator\product\11.2.0\dbhome_ 1\bin\oci... Make sure you have the 32

338. Counting Bits 比特位计数

https://leetcode-cn.com/problems/counting-bits/description/ Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary represent

The necessary bits to build these optional modules were not found: _uuid _bz2 _curse _curses_panel

在安装Python3.7可能遇到如题的错误，只需安装uuid库就可以 ubuntu下安装uuid链接库 sudo apt-get install uuid-dev CentOS 　　　　yum install libuuid-devel 对于以下的问题 The necessary bits to build these optional modules were not found: _bz2

[LeetCode] Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11’ has binary representation 0000000000

hdu 5106 Bits Problem(数位dp)

题目链接：hdu 5106 Bits Problem 题目大意：给定n和r，要求算出[0,r)之间所有n-onebit数的和。解题思路：数位dp，一个ct表示个数，dp表示和，然后就剩下普通的数位dp了。不过貌似正解是o(n)的算法，但是n才 1000，用o(n^2)的复杂度也是够的。 #include <cstdio>#include <cstring>#include <

UVA 11645 - Bits(数论+计数问题)

题目链接：11645 - Bits 题意：给定一个数字n，要求0-n的二进制形式下，连续11的个数。思路：和 UVA 11038 这题类似，枚举中间，然后处理两边的情况。不过本题最大的答案会超过longlong，要用高精度，不过借鉴http://www.cnblogs.com/TO-Asia/p/3214706.html这个人的方法，直接用两个数字来保存一个数字，这样能保

**Leetcode 190. Reverse Bits

for 8 bit binary number abcdefgh, the process is as follow: abcdefgh -> efghabcd -> ghefcdab -> hgfedcba 非常牛逼的写法。。转自： https://leetcode.com/problems/reverse-bits/discuss/54741/O(1)-bit-operatio

[python]使用bits表示weekdays

在做表设计的时候，可能会遇到需要表示Weekdays的场景： {id:0, text:"Mon"},{id:1, text:"Tue"},{id:2, text

[leetcode]--338. Counting Bits

Question 338. Counting Bits Question: Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them

《leetCode》：Number of 1 Bits

题目 Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).For example, the 32-bit integer ’11' has binary representation 00000000

《leetCode》： Reverse Bits

题目 Reverse bits of a given 32 bits unsigned integer.For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 001110010

网课：第二章模拟、枚举与贪心---Bits

题目 n组询问，每次给出一个区间l, r，你需要输出在这个区间内二进制表示中1的个数最多的数如有多个答案，输出最小的那个输入第一行包含整数 n -查询次数( 1 ≤ n ≤ 10000 )。下面每行 n 包含两个整数 li, ri --相应查询的参数( 0 ≤ li ≤ ri ≤ 1e18 )。输出对于每个查询，将答案打印在单独一行中。做法肯定不能for循环暴

【leetcode】经典算法题-Counting Bits

题目描述：给定一个数字n，统计0～n之间的数字二进制的1的个数，并用数组输出例子： For num = 5 you should return [0,1,1,2,1,2]. 要求：算法复杂复o（n）空间复杂度o（n）原文描述： Given a non negative integer number num. For every numbers i in the range 0

LeetCode--461. Hamming Distance 191. Number of 1 Bits 477. Total Hamming Distance

问题链接： https://leetcode.com/problems/hamming-distance/ https://leetcode.com/problems/number-of-1-bits/ https://leetcode.com/problems/total-hamming-distance/ 这三个都是有关hamming距离的问题，都是基于位运算相对简单基础的问题。问

Codeforces1208 F. Bits And Pieces (SOS DP)

题意：寻找最大的a[i]|a[j]&a[k]，满足i<j<k。思路： SOS DP。定义 d p [ s t a ] [ 0 / 1 ] dp[sta][0/1] dp[sta][0/1]得到 s t a sta sta这个状态的最大下标，次大下标。那么初始有 d p [ a [ i ] ] [ 0 ] = i dp[a[i]][0]=i dp[a[i]][0]=i。对于a[j]&a[

习题 8-3 比特变换器（Bits Equalizer, SWERC 2012, UVa12545）

原题链接：https://vjudge.net/problem/UVA-12545 分类：贪心法备注：简单思维题先看还少多少个’1’，如果s的’1’多余t的‘1’肯定无解。然后把少掉的这些’1’，全部用’?‘去变成’1’，如果’?‘有剩余则变“0‘。如果’?‘不够，则要把遍历一遍，当’1’的数量处于不足状态时，把没匹配的’0’变’1’。最后会剩下2*k个位置不匹配的字符，两两交换，贡献

systemverilog中for循环一个多bits variable的严格正确写法与综合比较

1.首先看某大厂给出的一小段代码：这段code什么意思呢？起初： 1）我认为是wvalid = 0； 2）下面循环的任何一次为1，就把wvalid赋值成1； 3）结果就是一旦wvalid为1，那么就永远为1，不管循环了多少次。 4）但是仔细想想，怎么会有这么vagary的逻辑呢？一定不是这样的，百思不得骑姐啊！！！ 5）只好综合一下了，看下图：原来是把

生信软件15 - 生信NGS数据分析强大的工具集ngs-bits

ngs-bits - Short-read sequencing tools - 短reasd测序工具一个强大的工具集，安装后直接使用。 # conda安装指定版本，安装高版本失败，故采用2021_09版本conda install ngs-bits=2021_09 -y# 测试安装是否成功，成功则显示下列内容ReadQC 软件说明文档markown # 参考githubh

LeetCode 题解：338. Counting Bits

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array. Example: For num = 5 you

vxWorks6.9动态加载程序时出现 Relocation value does not fit in 24/26 bits

在vxworks中动态加载可执行程序，有时候会出现类似 Relocation value does not fit in 26 bits (offset: 0x10, type: 1) 的错误提示，从而导致程序加载失败。出错原因： Some architectures have instructions that use less than 32 bits to reference a near

LeetCode-Number_of_1_Bits

题目： Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11' has binary representation 00000

LeetCode-Revers_Bits

题目： Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 001110

LeetCode 338 Counting Bits (递推)

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you