bitset

Given a certain integer x , I wish to compute the next higher integer y which has a certain hamming weight w . Mind that the hamming weight of x does not have to be w as well. So, for example x = 10 (1010) and w = 4 the result should be y = 15 (1111). Obviously, I could achieve this by just incrementing x but that would be a very slow solution for high numbers. Can I achieve this by the means of bit shifts somehow? There are three cases: the Hamming weight (a.k.a. bitwise population count) is reduced, unchanged, or increased. Increased Hamming weight Set enough successive low-order 0-bits to

I'm working on a structure that uses std::bitset and it looks like this: Register.h #pragma once #include <bitset> #include <vector> // used for typedefs of int types namespace vpc { // virtual pc typedef std::int8_t i8; typedef std::int16_t i16; typedef std::int32_t i32; typedef std::int64_t i64; const unsigned int BYTE = 0x08; const unsigned int WORD = 0x10; const unsigned int DWORD = 0x20; const unsigned int QWORD = 0x40; typedef std::bitset<BYTE> Byte; typedef std::bitset<WORD> Word; typedef std::bitset<DWORD> DWord; typedef std::bitset<QWORD> QWord; template<std::uint64_t N = BYTE> struct

Persistent Bookcase 题目链接 ： http://codeforces.com/contest/707/problem/D 注释 ：略。 题解 ： 发现虽然$q\le 10^5$但是网格是$1000\times 1000$的，而且每次操作只会操作一行。 故此我们考虑按照行来搞。 想到每次暴力重新建一行，但是空间开不下，我们用$bitset$即可。 但是我们又面临一个问题，即：回到某一个时刻。 这个很难弄，最简单的支持可持久化的数据结构是主席树，所以我们对行建主席树。 每次修改操作我们都新开一个$bitset$，主席树的第$i$个叶子表示的是第$i$行对应哪一个$bitset$。 时间复杂度为$O(\frac{np}{32} +plogn)$。 代码 ： #include <bits/stdc++.h> #define setIO(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout) #define N 400010 using namespace std; bitset <1010 > b[N], mdl; char *p1, *p2, buf[100000]; #define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000,

I have multiple threads in my application accessing a BitSet concurrently. The documentation says: A BitSet is not safe for multithreaded use without external synchronization. It doesn't say if it is not safe for reading or writing. Can anybody explain. A BitSet is only safe for read-only operations if there is a "happens before" relationship between the last action that initializes the BitSet and the actions that read it. The simplest way to achieve this is using a final . For example: public class BitsetHolder { private final BitSet b; public BitSetHolder() { b = new BitSet(); // operations

I can't figure out if I'm doing anything wrong here, hopefully someone here can enlighten me. I have a class Flags , this is an extremely simplified version but I declare a bitset class Flags { private: //List of 8 bits std::bitset<8> _P; public: Flags(); } On my constructor I initialise it as Flags::Flags() : _P(32ul) {} But it won't compile and gives me the error error C2668: 'std::bitset<_Bits>::bitset' : ambiguous call to overloaded function This is compiled in VS2010 SP1 64 bit but as a 32bit program EDIT Accepted answer is for the above but as a side note could anyone explain why when

I want to make a simple program that will take number of bits from the input and as an output show binary numbers, written on given bits (example: I type 3: it shows 000, 001, 010, 011, 100, 101, 110, 111) . The only problem I get is in the second for -loop, when I try to assign variable in bitset< bits > , but it wants constant number. If you could help me find the solution I would be really greatful. Here's the code: #include <iostream> #include <bitset> #include <cmath> using namespace std; int main() { int maximum_value = 0,x_temp=10; //cin >> x_temp; int const bits = x_temp; for (int i =