bitset | 易学教程

java.util.BitSet — set() doesn't work as expected

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问题 Am I missing something painfully obvious? Or does just nobody in the world actually use java.util.BitSet? The following test fails: @Test public void testBitSet() throws Exception { BitSet b = new BitSet(); b.set(0, true); b.set(1, false); assertEquals(2, b.length()); } It's really unclear to me why I don't end up with a BitSet of length 2 and the value 10. I peeked at the source for java.util.BitSet, and on casual inspection it seems to fail to make sufficient distinction between a bit that

Binary Serialization of std::bitset

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问题 std::bitset has a to_string() method for serializing as a char -based string of 1 s and 0 s. Obviously, this uses a single 8 bit char for each bit in the bitset, making the serialized representation 8 times longer than necessary. I want to store the bitset in a binary representation to save space. The to_ulong() method is relevant only when there are less than 32 bits in my bitset. I have hundreds. I'm not sure I want to use memcpy() / std::copy() on the object (address) itself, as that

Working with bitcode

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问题 This is slightly OT for SO, because I'm not trying to solve a specific problem, instead just to understand how something might be implemented. But I am after code, so let's see how it goes... Let's say we had a checkbox for each day of the week, and we decided to store any combination of those checkboxes as a single number, such that: 0 = no days 1 = Monday 2 = Tuesday 4 = Wednesday 8 = Thursday 16 = Friday 32 = Saturday 64 = Sunday 127 = everyday How might one go about implementing that

What is the size of bitset in C++

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问题 I want to know how bitset actually allocates memory. I read from some blog that it takes up memory in bits. However when i run the following code: bitset<3> bits = 001; cout<<sizeof(bits); I get the output as 4. What is the explanation behind it? Also is there a method to allocate space in bits in C++? 回答1: You can approximate sizeof(bitset<N>) as: If internal representation is 32bit (like unsigned on 32bit systems) as 4 * ((N + 31) / 32) If internal representation is 64bit (like unsigned

BitSet to and from integer/long

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问题 If I have an integer that I'd like to perform bit manipulation on, how can I load it into a java.util.BitSet ? How can I convert it back to an int or long? I'm not so concerned about the size of the BitSet -- it will always be 32 or 64 bits long. I'd just like to use the set() , clear() , nextSetBit() , and nextClearBit() methods rather than bitwise operators, but I can't find an easy way to initialize a bit set with a numeric type. 回答1: The following code creates a bit set from a long value

bitset用法和ch2101

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我又回来啦第一次尝试使用markdown文本，可能编辑不好，见谅咕咕咕题目 (这个网站里有全部蓝书的题目， good ) 咳咳下面步入正题，关于这种题目，看到的第一眼应该能想到相当于动规的递推，但是要求边界如果直接dfs，30000的数据不知道你家的栈能不能~受~的住所以这个题目提供了一个很好的思路，就是先求出拓扑序然后反过来一个一个求，这样可以保证求当前节点数据时（小于等于30000不用longlong）所用到的节点都以经算好了，就这么简单就这么简单？？你好好考虑过动归方程了吗？ f(x)={x}并(所有f(y)的并,存在有向边(x,y)) 为甚不是加？？因为有可能会有重复。比如3->4,2->4那直接加起来，4就会算两次怎么破？？主角闪亮登场！！状压用N位二进制数，反之一个点到其他点只有能与不能 1就是能呗，2就是不能呗实现：数组实现（不够装逼） bitset! 一直不知道bitset怎么用今天先码几个呗（~嘤嘤嘤~） 1：定义 bitset名称 bitset名称[数组大小] 2:运算假设有bit[30001] 可以直接赋值bit[1][1]=1; 就是第一个1位为1 可以直接bit[1]|bit[2] (全部按位或) 先到这里，走了，下次再补吧 19.12.15 20.14 来源： https://www.cnblogs.com/Q-M

What is the C# equivalent of BitSet of Java?

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问题 My requirements are simple: Be able to define dimensions of bit array, i.e: 5 bytes. bool Get(bitIndex: int) Set(bitIndex: int) Is there a c# equivalent which provides similar functionality to BitSet in Java? Here's the scenario: Initialize 5 bytes, all bits are 0(false). Set byte 3, bit 8 to TRUE. Get status of byte 3, bit 8. UPDATE : Solution from Michael Bray: static void Main(string[] args) { // Set for 5 bytes BitArray ba = new BitArray(8 * 5); // Set bit #1 on byte #4 ba.Set(GetBitNum(4

Performance of doing bitwise operations on bitsets

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问题 In C++ if I do a logical OR (or AND) on two bitsets, for example: bitset<1000000> b1, b2; //some stuff b1 |= b2; Does this happen in O(n) or O(1) time? Why? Also, can this be accomplished using an array of bools in O(1) time? Thanks. 回答1: It has to happen in O(N) time since there is a finite number of bits that can be processed in any given chunk of time by a given processor platform. In other words, the larger the bit-set, the longer the amount of time each operation will take, and the

Data structure recommendation

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问题 Developing in Java, I need a data structure to select N distinct random numbers between 0 and 999999 ? I want to be able to quickly allocate N numbers and make sure they don't repeat themselves. Main goal is not to use too much memory and still keep performance reasonable. I am considering using a BitSet But I am not sure if the memory implications. Can someone tell me if the memory requirements of this class are related to the number of bits or to the number of set bits? and what is the

How do I get the number of bits represented by a BitSet?

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问题 If I run the following code: BitSet test = new BitSet(20); System.out.println("Bitset size is " + test.size()); System.out.println("Bitset length is " + test.length()); I get the output: Bitset size is 64 Bitset length is 0 Which makes sense now that I look closer at the documentation (the former getting the size including implementation overhead and the latter being the last set bit, with all defaulting to false), but is not what I want. Since the BitSet I actually use can have varying