blocks|key|2529926|text|你只需要计算每一种情况的范围，找出2的最低幂，它比那个范围要高。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2529927|例如，在i中，3位小数->+10%5E3+=1000个可能的数字，因此您必须找到高于1000的最小幂2，在本例中是2%5E10+=+1024+(10位)。|2529928|编辑：--基本上，您需要用您拥有的数字数来查找可能的数字数，然后找出哪些数字(在另一个基中，在本例中是2，二进制)至少具有与十进制中的数字相同的可能数。|offset|length|style|BOLD|2529929|若要计算给定数字数的可能性数：possibilities=base%5Endigits|CODE|2529930|因此，如果小数位数为3位(基数为10)，则有10%5E3=1000可能性。然后，您必须在二进制(位，基2)中找到一些数字，以便可能性的数量至少为1000，在本例中为2%5E10=1024+(9位数字不够，因为2%5E9=512小于1000)。|2529931|如果你概括了这一点，你就有：2%5Enbits=possibilities+<=>+nbits=log2(possibilities)|2529932|它适用于i)，给出：log2(1000)=9.97，由于位数必须是整数，所以必须将其舍入到10。|2529933|entityMap^0|0|0|0|3|0|F|Q|0|M|9|28|9|2S|7|0|E|1F|0|A|F|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@$F|X|G|Y|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|Z|8|@$F|10|G|11|H|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|12|8|@$F|13|G|14|H|L]|$F|15|G|16|H|L]|$F|17|G|18|H|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|19|8|@$F|1A|G|1B|H|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1C|8|@$F|1D|G|1E|H|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|T|$]]

Well, you just have to calculate the range for each case and find the lowest power of 2 that is higher than that range.

For instance, in i), 3 decimal digits -> 10^3 = 1000 possible numbers so you have to find the lowest power of 2 that is higher than 1000, which in this case is 2^10 = 1024 (10 bits).

Edit: Basically you need to find the number of possible numbers with the number of digits you have and then find which number of digits (in the other base, in this case base 2, binary) has at least the same possible numbers as the one in decimal.

To calculate the number of possibilities given the number of digits: <code>possibilities=base^ndigits</code>

So, if you have 3 digits in decimal (base 10) you have <code>10^3=1000</code> possibilities. Then you have to find a number of digits in binary (bits, base 2) so that the number of possibilities is at least 1000, which in this case is <code>2^10=1024</code> (9 digits isn't enough because <code>2^9=512</code> which is less than 1000).

If you generalize this, you have: <code>2^nbits=possibilities &lt;=&gt; nbits=log2(possibilities)</code>

Which applied to i) gives: <code>log2(1000)=9.97</code> and since the number of bits has to be an integer, you have to round it up to 10.

blocks|key|761079|text|把这个数字除以2，直到你得到一个0的商数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|761080|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Keep dividing the number by 2 until you get a quotient of 0.

blocks|key|2530006|text|最简单的答案是将所需的值转换为二进制，并查看该值所需的位数。然而，这个问题询问一个十进制数的X位数有多少位。在这种情况下，您似乎必须选择X位数的最高值，然后将该数字转换为二进制数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2530007|作为一个基本的例子，让我们假设我们想存储一个1位的基数为10的数字，并且想知道需要多少位。最大的1位基数是9，所以我们需要把它转换成二进制。这产生1001，总共有4位。这个示例可以应用于一个两位数的数字(最大值为99，转换为1100011)。要对n个数字进行求解，可能需要对其他数字进行求解并搜索模式。|2530008|若要将值转换为二进制值，请重复将值除以二，直到得到商数为0(所有剩馀数都将为0或1)。然后你倒转你的余数的顺序，得到二进制的数字。|2530009|例子:+13到二进制。|2530010|1|offset|length|style|BOLD|2530011|6/2+=3r+|unordered-list-item|2530012|
|2530013|3/2+=1r+1|2530014|2530015|1/2+=0+r+1|2530016|2530017|=+1101+(8*1)%2B+(4*1)+%2B+(2*0)+%2B(1*1)|2530018|2530019|希望这能帮上忙。|2530020|entityMap^0|0|0|0|0|0|1|0|0|0|8|1|0|0|9|1|0|0|2|4|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|17|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|18|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|19|8|@$J|1A|K|1B|L|M]]|9|@]|A|$]]|$1|N|3|O|5|P|7|1C|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1D|8|@]|9|@]|A|$]]|$1|S|3|T|5|P|7|1E|8|@$J|1F|K|1G|L|M]]|9|@]|A|$]]|$1|U|3|R|5|6|7|1H|8|@]|9|@]|A|$]]|$1|V|3|W|5|P|7|1I|8|@$J|1J|K|1K|L|M]]|9|@]|A|$]]|$1|X|3|R|5|6|7|1L|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|P|7|1M|8|@$J|1N|K|1O|L|M]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]]|14|$]]

The simplest answer would be to convert the required values to binary, and see how many bits are required for that value. However, the question asks how many bits for a decimal number of X digits. In this case, it seems like you have to choose the highest value with X digits, and then convert that number to binary. 

As a basic example, Let's assume we wanted to store a 1 digit base ten number, and wanted to know how many bits that would require. The largest 1 digit base ten number is 9, so we need to convert it to binary. This yields 1001, which has a total of 4 bits. This same example can be applied to a two digit number (with the max value being 99, which converts to 1100011). To solve for n digits, you probably need to solve the others and search for a pattern.

To convert values to binary, you repeatedly divide by two until you get a quotient of 0 (and all of your remainders will be 0 or 1). You then reverse the orders of your remainders to get the number in binary. 

Exampe: 13 to binary.

<ul>
<li>13/2 = 6 r 1</li>
<li>6/2 = 3 r 0</li>
<li>3/2 = 1 r 1</li>
<li>1/2 = 0 r 1</li>
<li>= 1101 ((8*1) + (4*1) + (2*0) + (1*1))</li>
</ul>

Hope this helps out.

blocks|key|2549331|text|为了概括表示一个数字需要多少位的技术，就这样做了。你有一个表示的R符号，你想知道有多少位，解这个方程R=2%5En或log2(R)=n，其中n是位数，R是表示符号的数目。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2549332|对于十进制数字系统R=9，所以我们求解9=2%5En，答案是每小数位3.17位。因此，一个3位数字需要9.51位或10位。一个1000位数字需要3170位|2549333|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|$]]

Ok to generalize the technique of how many bits you need to represent a number is done this way. You have R symbols for a representation and you want to know how many bits, solve this equation R=2^n or log2(R)=n. Where n is the numbers of bits and R is the number of symbols for the representation. 

For the decimal number system R=9 so we solve 9=2^n, the answer is 3.17 bits per decimal digit. Thus a 3 digit number will need 9.51 bits or 10. A 1000 digit number needs 3170 bits

blocks|key|2549376|text|让它所需的n位，然后是2%5En=(基)%5E位，然后取日志并计数no。为n|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2549377|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

let its required n bit then 2^n=(base)^digit and then take log and count no. for n

blocks|key|2530155|text|对于n位数的二进制数，它可以容纳的最大十进制值为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2530156|2%5En+-+1和2%5En是可以使用这几个数字生成的总排列。|2530157|以你只想要三位数的情况为例(你的案例1。我们看到要求是，|2530158|2%5En+-1+>=+999|2530159|向双方申请日志，|2530160|日志(2%5En-1)+>=日志(999)|2530161|log(2%5En)+-+log(1)+>=+log(999)|2530162|N=+9.964+(约)。|2530163|取9.964以来n的ceil值不能是一个有效的数字数，我们得到n=+10。|2530164|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|Z|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|10|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|11|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|S|$]]

For a binary number of n digits the maximum decimal value it can hold will be

2^n - 1, and 2^n is the total permutations that can be generated using these many digits.

Taking a case where you only want three digits, ie your case 1. We see that the requirements is,

2^n - 1 >= 999

Applying log to both sides,

log(2^n - 1) >= log(999)

log(2^n) - log(1) >= log(999)

n = 9.964 (approx).

Taking the ceil value of n since 9.964 can't be a valid number of digits, we get n = 10.

blocks|key|757907|text|假设问题是问你要存储的最小位数是多少|type|unstyled|depth|inlineStyleRanges|entityRanges|data|757908|757909|3位数|ordered-list-item|757910|757911|我处理这个问题的方法是：|757912|757913|我们需要存储的最大3位数是多少？Ans:+999|unordered-list-item|757914|存储这个数字所需的最小位数是多少？|757915|757916|这个问题可以通过递归地除以999除以2来解决。然而，用数学的力量来帮助我们就更简单了。从本质上讲，我们正在为下面的方程求解n：|757917|2%5En+=+999
nlog2+=+log999
n+~+10|code-block|syntax|javascript|757918|你需要10位才能存储3位数字。|757919|使用类似的方法来解决其他的子问题！|757920|希望这能有所帮助！|757921|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|16|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|18|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|1A|8|@]|9|@]|A|$]]|$1|M|3|N|5|L|7|1B|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1D|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|1E|8|@]|9|@]|A|$U|V]]|$1|W|3|X|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1G|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1H|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|13|$]]

Assuming that the question is asking what's the minimum bits required for you to store 

<ol>
<li>3 digits number</li>
</ol>

My approach to this question would be: 

<ul>
<li>what's the maximum number of 3 digits number we need to store? Ans: 999</li>
<li>what's the minimum amount of bits required for me to store this number?</li>
</ul>

This problem can be solved this way by dividing 999 by 2 recursively. However, it's simpler to use the power of maths to help us. Essentially, we're solving n for the equation below:

<pre><code>2^n = 999
nlog2 = log999
n ~ 10
</code></pre>

You'll need 10 bits to store 3 digit number. 

Use similar approach to solve the other subquestions!

Hope this helps!

blocks|key|2530230|text|这里有很多答案，但我会添加我的方法，因为我发现这篇文章是在处理同样的问题时发现的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2530231|从我们所知道的数字开始，从0到16。|2530232|Number+++++++++++encoded+in+bits+++++++++minimum+number+of+bits+to+encode
0++++++++++++++++000000++++++++++++++++++1
1++++++++++++++++000001++++++++++++++++++1
2++++++++++++++++000010++++++++++++++++++2
3++++++++++++++++000011++++++++++++++++++2
4++++++++++++++++000100++++++++++++++++++3
5++++++++++++++++000101++++++++++++++++++3
6++++++++++++++++000110++++++++++++++++++3
7++++++++++++++++000111++++++++++++++++++3
8++++++++++++++++001000++++++++++++++++++4
9++++++++++++++++001001++++++++++++++++++4
10+++++++++++++++001010++++++++++++++++++4
11+++++++++++++++001011++++++++++++++++++4
12+++++++++++++++001100++++++++++++++++++4
13+++++++++++++++001101++++++++++++++++++4
14+++++++++++++++001110++++++++++++++++++4
15+++++++++++++++001111++++++++++++++++++4
16+++++++++++++++010000++++++++++++++++++5|code-block|syntax|javascript|2530233|看看休息时间，它显示了这张桌子。|2530234|number+<=+++++++number+of+bits
1+++++++++++++++0
3+++++++++++++++2
7+++++++++++++++3
15++++++++++++++4|2530235|那么，现在我们如何计算模式呢？|2530236|记住，原木基2+(n)+=原木基10+(n)+/原木基10+(2)|2530237|number++++logb10+(n)+++logb2+(n)+++ceil[logb2(n)]+
1+++++++++0++++++++++++0+++++++++++0+++++++++++(special+case)
3+++++++++0.477++++++++1.58++++++++2
7+++++++++0.845++++++++2.807+++++++3++
8+++++++++0.903++++++++3+++++++++++3+++++++++++(special+case)
15++++++++1.176++++++++3.91++++++++4
16++++++++1.204++++++++4+++++++++++4+++++++++++(special+case)
31++++++++1.491++++++++4.95++++++++5
63++++++++1.799++++++++5.98++++++++6|2530238|现在，所需的结果与第一个表匹配。注意一些值也是特例。当你应用上限的时候，这些值不会改变，所以你知道你需要加1才能得到最小的位字段长度。|2530239|若要说明特殊情况，请在输入中添加一个。在python中实现的结果代码是：|2530240|from+math+import+log
from+math+import+ceil
def+min_num_bits_to_encode_number(a_number):
++++a_number=a_number%2B1++#+adjust+by+1+for+special+cases

++++#+log+of+zero+is+undefined
++++if+0==a_number:
++++++++return+0
++++num_bits+=+int(ceil(log(a_number,2)))++#+logbase2+is+available
++++return+(num_bits)|2530241|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

There are a lot of answers here, but I'll add my approach since I found this post while working on the same problem.

Starting with what we know here are the number from 0 to 16.

<pre><code>Number encoded in bits minimum number of bits to encode
0 000000 1
1 000001 1
2 000010 2
3 000011 2
4 000100 3
5 000101 3
6 000110 3
7 000111 3
8 001000 4
9 001001 4
10 001010 4
11 001011 4
12 001100 4
13 001101 4
14 001110 4
15 001111 4
16 010000 5
</code></pre>

looking at the breaks, it shows this table

<pre><code>number &lt;= number of bits
1 0
3 2
7 3
15 4
</code></pre>

So, now how do we compute the pattern?

Remember that log base 2 (n) = log base 10 (n) / log base 10 (2)

<pre><code>number logb10 (n) logb2 (n) ceil[logb2(n)] 
1 0 0 0 (special case)
3 0.477 1.58 2
7 0.845 2.807 3 
8 0.903 3 3 (special case)
15 1.176 3.91 4
16 1.204 4 4 (special case)
31 1.491 4.95 5
63 1.799 5.98 6
</code></pre>

Now the desired result matching the first table. Notice how also some values are special cases. 0 and any number which is a powers of 2. These values dont change when you apply ceiling so you know you need to add 1 to get
the minimum bit field length. 

To account for the special cases add one to the input. The resulting code implemented in python is:

<pre><code>from math import log
from math import ceil
def min_num_bits_to_encode_number(a_number):
 a_number=a_number+1 # adjust by 1 for special cases

 # log of zero is undefined
 if 0==a_number:
 return 0
 num_bits = int(ceil(log(a_number,2))) # logbase2 is available
 return (num_bits)
</code></pre>

blocks|key|761220|text|这个能用！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|761221|floor(loge(n)+/+loge(2))+%2B+1|code-block|syntax|javascript|761222|要包含负数，您可以添加一个额外的位来指定符号。|761223|floor(loge(abs(n))+/+loge(2))+%2B+2|761224|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This one works!

<pre><code>floor(loge(n) / loge(2)) + 1
</code></pre>

To include negative numbers, you can add an extra bit to specify the sign.

<pre><code>floor(loge(abs(n)) / loge(2)) + 2
</code></pre>

blocks|key|2530259|text|简单的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2530260|int+nBits+=+ceil(log2(N));|code-block|syntax|javascript|2530261|这只是因为pow(2，nBits)略大于N。|offset|length|style|BOLD|2530262|entityMap^0|0|0|5|C|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The short answer is:

<pre><code>int nBits = ceil(log2(N));
</code></pre>

That's simply because pow(2, nBits) is slightly bigger than N.

This is a homework question that I am stuck with.

<blockquote>
 Consider unsigned integer representation. How many bits will be
 required to store a decimal number containing:
</blockquote>

<pre><code>i) 3 digits ii) 4 digits iii) 6 digits iv) n digits
</code></pre>

I know that the range of the unsigned integer will be 0 to 2^n but I don't get how the number of bits required to represent a number depends upon it. Please help me out.

Thanks in advance.

Calculating bits required to store decimal number

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

这是一个家庭作业问题，我被困住了。考虑无符号整数表示。存储包含以下内容的十进制数需要多少位：i) 3 digits ii) 4 digits iii) 6 digits iv) n digits我知道无符号整数的范围是0到2^n，但我不知道表示一个数字所需的位数如何取决于它。请帮帮我。提前谢谢。

问计算存储十进制数所需的位数
EN

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

问计算存储十进制数所需的位数EN