blocks|key|1493445|text|你有vec1，它是一辆std::vector<T>，由这辆卡车代表，它的拖车上有一个黑色集装箱。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1493446|​|1493447|📷|atomic|1493448|1493449|(资料来源:+bigcommerce.com)|1493450|现在，它的黑容器里有数以百万计的文件(用黑色容器，std::vector<T>对元素的记忆)。现在，你想把这些文件复制到另一种类似的工具，vec2。你怎样才能做到最好?照片-复制的复杂性是什么?它将是O(论文的数量)。|BOLD|1493451|现在，假设你想移动的内容.std::vector所需要做的就是把它的黑色容器从卡车上分离出来，然后把它连接到另一辆卡车上.这使得将容器的复杂性移动为O(1)。|1493452|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/7bc20d43a1129ce7c12e46d632b845bb.jpg|imageAlt^0|2|4|B|E|0|0|0|1|0|0|0|0|P|E|1X|4|2F|2|0|7|2|1S|4|D|B|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|17|8|@]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|K|3|G|5|6|7|1B|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1C|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|P]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1K|8|@$9|1L|A|1M|B|P]|$9|1N|A|1O|B|P]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z|10|-4]]]]

You have <code>vec1</code> which is a <code>std::vector&lt;T&gt;</code> represented by this truck with a black-container attached to its trailer. 

<a href="https://i.stack.imgur.com/0tece.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0tece.jpg" alt="enter image description here"></a>

(source: bigcommerce.com)

Now, there are millions of papers in its black-container (by black-container, <code>std::vector&lt;T&gt;</code>'s memory for the elements). Now, you want to photocopy the papers into another similar vehicle, <code>vec2</code>. How best can you do it? What is the photo-copy complexity? It is going to be O(Number of papers)

<hr>

Now, assuming you want to move the contents... All <code>std::vector</code> needs to do is to detach its black-container from the truck, and attach it into another truck's... That makes move complexity of the container to be O(1)

blocks|key|198188|text|要理解操作的复杂性，首先必须定义计算模型。特别是，您必须定义哪些操作是O(1)，哪些操作是免费的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|198189|例如，如果您正在查看需要大量硬盘访问的程序，您将只考虑将一个块从磁盘读取为一个操作，而将内存访问视为空闲。|198190|在位掩码和位集的示例中，您可能正在计算内存块复制操作。然后，复制一个适合于一个内存块的位掩码只需要一个内存副本。但是，复制位掩码在这个模型中仍然是O(N)，因为一个大位掩码将跨越许多内存块。更准确地说，如果b是一个内存块中的位数，N是位掩码中的位数，那么您需要ceil(N/b)内存块复制操作来复制位掩码。|198191|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

To understand the complexity of an operation, you first have to define your model of computation. In particular, you have to define what operations are O(1) and what operations you consider free. 

For example, if you are looking at programs needing a lot of hard disk access, you would only consider reading a block from disk as one operation and memory access as free. 

In the example of bitmasks and bitssets, you are probably counting memory block copy operations. Then copying a bit mask that fits into one memory block costs only one memory copy. However, copying bitmasks is still O(N) in this model because a large bitmask will span many memory blocks. More precisely, if b is the number of bits in one block of memory and N the number of bits in the bitmask, then you need ceil(N/b) memory block copy operations to copy the bitmask.

blocks|key|1973040|text|复制长度为N的向量可以通过以下几种方法来测量：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1973041|O(1)向量复制操作|unordered-list-item|1973042|O(N)元素复制操作|1973043|O(NT)度量单位，如果元素复制操作为O(T)|1973044|顺便说一句，你的朋友对bitset的看法是错误的，或者你误解了。bitset的优点是对数据结构进行了优化，以利用可以由单个线程执行的低级别并行执行形式--即64位整数操作可以同时在64位上工作。|1973045|entityMap^0|5|1|0|0|4|0|0|4|0|0|5|J|4|0|B|6|W|6|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|I|3|J|5|H|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|K|3|L|5|H|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|P|$]]

Copying a vector of length <code>N</code> can be measured in several ways:

<ul>
<li><code>O(1)</code> vector copy operations</li>
<li><code>O(N)</code> element copy operations</li>
<li><code>O(NT)</code> units of measure, if the element copy operation is <code>O(T)</code></li>
</ul>

Incidentally, your friend is wrong about <code>bitset</code>, or you misunderstood. The advantage of <code>bitset</code> is that the data structure is optimized to take advantage of a low level form of parallel execution that can be done by a single thread &mdash; i.e. 64-bit integer operations can be used to work on 64 different bits simultaneously.

A friend of mine said that bitmasks are better than bitsets because it takes O(N) time to copy a bitset whereas only O(1) time to copy a bitmask, making it far faster for applications such as in Dynamic Programming. So I was wondering is it the same for all other data types?

For eg what would be the complexity of this block of code?

<pre><code>vector &lt;int&gt; vec1 = {1,2,3,4,5,6,7,8,9,10};
vector &lt;int&gt; vec2 = vec1;
</code></pre>

If it is O(N), why is it so? Shouldn't it be O(1) just like doing something like

<pre><code>int x = 530210;
int y = x;
</code></pre>

would be?

What is the complexity of copying a variable in C++

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我的一位朋友说，位掩码比位集更好，因为复制位集需要O(N)时间，而复制位掩码只需O(1)时间，这样对于动态编程之类的应用程序来说要快得多。所以我想知道所有其他数据类型都是一样的吗？例如，这段代码的复杂性是什么？vector <int> vec1 = {1,2,3,4,5,6,7,8,9,10};vector <int>...

问在C++中复制变量的复杂性是多少？
EN

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问在C++中复制变量的复杂性是多少？EN