bitset

%%%外校参加国赛大佬kai神 220 -rank1 ---------------以上选手实力开挂---------------- %%%skyh、NC锅 180 -rank2 %%%神牛170-rank4 %%%迪哥、有钱人160-rank5 话说我刚颓完bitset就考这回事…… 看来我bitset还是没颓全啊QAQ…… 挂成41名了…… T1想的时间太长了。一直以为有什么特殊性质可以优化。结果是一个bitset优化sb暴力…… 然后去看T2，一直觉得二分图可以拿的部分分挺足的结果二分图板子打挂了。 T3除了开始通读三道题的那几眼之外根本连看都没看。 还是押宝了。押的还不是AC。 赛后回宿舍睡前想到自己一开始想的那个答案统计方法完全可以让我的T1再多20分，然后就睡不着了。 退退退，退到无路可退。LGR你到底想要什么？ 来源： https://www.cnblogs.com/xingmi-weiyouni/p/11515989.html

好博客： https://www.cnblogs.com/cjjsb/p/9751384.html 例题1： 题目链接： https://www.nowcoder.com/acm/contest/132/C 题意： 分析：核心就在于看作01背包的形式，枚举种类n，每个范围（L-R+1），以及所有的可能值 1~1e6， 然后直接做的话又会超时，所以要用到 bitset 的位运算，将b[0][0]设置为1，每次向左移动 j*j 位，最后就可以得到经过所有组合后的可能取值。 #include <cstdio> #include <bitset> using namespace std; bitset<1000005>b[105]; int n,l[105],r[105]; int main() { scanf("%d",&n); int Min=0; for(int i=1;i<=n;i++) {scanf("%d%d",&l[i],&r[i]); Min+=l[i]*l[i];} b[0].set(0); //上面那句就等同于 -> b[0][0]=1; for(int i=1;i<=n;i++) for(int j=l[i];j<=r[i];j++) b[i]|=(b[i-1]<<(j*j)); // int ans=0; // for(int i=Min; i<=1000005;

How to write bitset data to a file? The first answer doesn't answer the question correctly, since it takes 8 times more space than it should. How would you do it ? I really need it to save a lot of true/false values. Simplest approach : take consecutive 8 boolean values, represent them as a single byte, write that byte to your file. That would save lot of space. In the beginning of file, you can write the number of boolean values you want to write to the file; that number will help while reading the bytes from file, and converting them back into boolean values! If you want the bitset class

In my program I need to check if I have already generated a value in a set of 2.5*10^9. I expect to generate about the half of the set and need to have a fast way to check and update it. The bitset seemed to me as a good idea as it takes not too much memory (1 bit per value) and is fast. The problem is that when I define my set in my class, I got a segmentation fault as the size is too big (it works with smaller sizes). private: std::bitset<2500000000UL> cover; // not working std::bitset<25000UL> cover; // working Any idea ? Thank you PS: I'd rather not use external library if possible. I'm

Is there a Python class or module that implements a structure that is similar to the BitSet? There's nothing in the standard library. Try: http://pypi.python.org/pypi/bitarray Emil Have a look at this implementation in Python 3. The implementation basically makes use of the built-in int type, which is arbitrary precision integer type in Python 3 (where long is the Python 2 equivalent). #! /usr/bin/env python3 """ bitset.py Written by Geremy Condra Licensed under GPLv3 Released 3 May 2009 This module provides a simple bitset implementation for Python. """ from collections import Sequence import

what is the best way to concatenate 2 bitsets? For example i've got boost::dynamic_bitset<> test1( std::string("1111") ); boost::dynamic_bitset<> test2( std::string("00") ); they should be concatenated into a thrid Bitset test3 which then holds 111100 Solutions should use boost::dynamic_bitset. If the solution works with std::bitset, it would be nice too. There should be a focus on performance when concatenating the bits. UPDATE: I've compared both methods (stringmethod from me and Neil and shiftmethod from messenger) and the stringmethod was a lot faster (factor 10++). Code here: http:/

In c++. I initialize a bitset to -3 like: std::bitset<32> mybit(-3); Is there a grace way that convert mybit to -3 . Beacause bitset object only have methods like to_ulong and to_string . Use to_ulong to convert it to unsigned long , then an ordinary cast to convert it to int . int mybit_int; mybit_int = (int)(mybit.to_ulong()); DEMO 来源： https://stackoverflow.com/questions/19583720/convert-bitset-to-int-in-c

Is there a way of iterating over a (possibly huge) std::bitset that is linear in the number of bits that are set to true ? I want to prevent having to check every single position in the bitset. The iteration should successively return the indices of each bit that is set to true. A standard bitvector does not support efficient iteration over true bits - the runtime is always O(n), where n is the number of total bits, which has no dependence on k. However, a special structure called a van Emde Boas Tree supports iteration over the bits in time O(k lg lg n), where n is the number of bits and k is