blocks|key|579020|text|99+999的二进制数是11000011010011111，就像你的计算器说的那样。我不知道你输入了什么，但很明显你在某个地方做了一个错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|579021|17位可以存储最多2%5E17的数字，因此=+131072，而16位只能存储最多2%5E16+=+65536。|579022|所以你至少需要17位来存储99+999，而不是更多，也不是更少。|579023|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|$]]

99 999 in binary is 11000011010011111, exactly as your calculator says. I don't know what you entered but clearly you made a typo somewhere.

17 bits can store a number up to 2^17, hence = 131072, while 16 bits can only store up to 
2^16 = 65536.

So you do need at least 17 bits to store 99 999, not any more, not any less.

blocks|key|240654|text|2%5E16+=+65,536，所以要捕获每5位数字，就需要2%5E17+=+17位。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|240655|关于“第一个二元值.”的部分对我来说毫无意义|240656|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

2^16 = 65,536, so to capture every 5 digit number, you'd need 2^17 = 17 bits.

The part about "first binary value ..." makes no sense to me

blocks|key|240688|text|嗯，16位可以存储一个与2%5E16+-+1+=+65535一样高的数字。这是假设无符号数字(非负)，这意味着它不足以表示5位数字的十进制数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|240689|使用17位，您可以得到与2%5E17+-1+=+131071一样高的无符号数字。|240690|entityMap^0|C|G|0|C|G|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

Well, 16 bits can store a number as high as <code>2^16 - 1 = 65535</code>. That's assuming unsigned numbers (non-negative), which means it is not enough to represent of 5 digits decimal numbers.

With 17 bits you can get represent unsigned numbers as high as <code>2^17 -1 = 131071</code>.

blocks|key|578995|text|我使用的不是在线计算器，而是使用Java和Integer.toBinaryString(int)，比如-|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|578996|int+val+=+99999;
String+binaryString+=+Integer.toBinaryString(val);
System.out.printf("%25d+is+binary+%25s+and+%25d+binary+digits+long",+val,
++++++++binaryString,+binaryString.length());|code-block|syntax|javascript|578997|输出是|578998|99999+is+binary+11000011010011111+and+17+binary+digits+long|578999|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toBinaryString%2528int%2529^0|L|R|L|R|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Rather then using an online calculator, I would use Java and <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toBinaryString%28int%29" rel="nofollow"><code>Integer.toBinaryString(int)</code></a> with something like -

<pre><code>int val = 99999;
String binaryString = Integer.toBinaryString(val);
System.out.printf("%d is binary %s and %d binary digits long", val,
 binaryString, binaryString.length());
</code></pre>

And the output is

<pre><code>99999 is binary 11000011010011111 and 17 binary digits long
</code></pre>

blocks|key|240827|text|考虑存储5位小数位数的可能性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|240828|00000
11111
22222
33333
44444
55555
66666
77777
88888
99999|code-block|syntax|javascript|240829|对于每个数字，您有10+(即“基数10")可能值，而对于5位数字，您有10*10*10*10*10+=+10%5E5+=+100000+(0是计数的)可能值。|240830|对于17位二进制数：|240831|00000000000000000
11111111111111111|240832|使用相同的逻辑，您有2*2*.*2=+2%5E17+=+131072的可能值，范围从0到131071。为什么你需要17位？嗯，2%5E16+=+65536，不能容纳99999。|240833|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Consider the possibilities for storing a 5 digit decimal number:

<pre><code>00000
11111
22222
33333
44444
55555
66666
77777
88888
99999
</code></pre>

For each digit, you have 10 (i.e. "base 10") possible values, and with 5 digits you have 10*10*10*10*10 = 10^5 = 100000 (0 is counted) possible values.

For a 17 digit binary number:

<pre><code>00000000000000000
11111111111111111
</code></pre>

Using the same logic, you have 2*2*...*2 = 2^17 = 131072 possible values ranging from 0 to 131071. As to why you need 17 bits? Well, 2^16 = 65536, which can't hold 99999.

I was reading this <a href="https://stackoverflow.com/questions/11179516/memory-required-to-store-1000-telephone-numbers">question</a> which stated it takes 17 bits to store 5 digits, how is that possible? Here is what I understand: 

I used the <a href="http://acc6.its.brooklyn.cuny.edu/~gurwitz/core5/nav2tool.html" rel="nofollow noreferrer">following calculator</a> and entered the number 99999. 

I got this huge number:

<code>11000000111001110000110100111111100001101001111111000011010011111</code>

From what I understand, each <code>0</code> and <code>1</code> represents one bit. As you can see, there are more than 17 bits!

Am I understanding this wrong?

Also, it was also stated:

<blockquote>
 "Now the first binary value which allows you to fit 99999 numbers is 131072,"
</blockquote>

Huh? I thought binary values are ONLY 0 and 1? What am I missing here?

new to programming - it takes 17 bits to store 5 digits - how?

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我读过这个，它说要用17位来存储5位数字，这怎么可能？以下是我的理解：我使用了并输入了数字99999。我得到了一个巨大的数字：11000000111001110000110100111111100001101001111111000011010011111据我所知，每个0和1都代表了一点。正如你所看到的，有超过17位！我是不是理解错了？此外，还指出：“现在，第一个允许你适应99999位数的二进制值

问新编程-它需要17位来存储5位数字-怎么做？
EN

回答 5

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问新编程-它需要17位来存储5位数字-怎么做？EN

回答 5

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问新编程-它需要17位来存储5位数字-怎么做？
EN