You have a positive integer try to find out how many digits it has

Count number of digits in a number in Python

Count the number of digits in a number using python :

Using python, count the number of digits in a number. In this tutorial, we will learn how to count the total number of digits in a number using python. The program will get the input from the user and print out the result.

We will show you two different ways to calculate total digits in a number.

Solution 1 (using a while loop) :

The idea behind this solution is to keep deleting the rightmost digit of the number one by one until the number becomes zero.

You can also download this program from here.

Solution 2 ( Using inbuilt methods ) :

Instead of using a while loop, we can directly convert the integer to a string and get the length of the string. In python, getting the length of a string is easy by using the ’len()’ method. For converting a number to a string, we can use the ’str()’ method.

‘abs’ is used to get the absolute value of a number. i.e. to handle the case even if the number is negative.

For example, if our input value is ’-123‘, abs() will convert it to ’123‘. str() will convert it to a string and len() will return the length of the string or 3.

You can also download this program from here.

You have a positive integer try to find out how many digits it has

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If m is a positive integer, then m^3 has how many digits?

(1) m has 3 digits.
(2) m^2 has 5 digits.

Most Helpful Expert Reply

If m is a positive integer, then m^3 has how many digits?

Pick some easy numbers.

(1)+(2) The same example worked for both statements so even taken together statements are not sufficient.

Answer: E.
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Most Helpful Community Reply

If m is a positive integer, then m^3 has how many digits?
1. m has 3 digits
2. m^2 has 5 digits

How would you do this quickly? Is there a rule that I am unaware of? I could do it, but I had to pick a few numbers. Thanks.

Back to original qtn:

If m is a positive integer, then \(m^3\) has how many digits?
stmnt1: \(m\) has 3 digits
==> \(10^2 \(10^6 \(m^3\) can have minimum of 7 (i.e 6+1) and max of 11 digits (i.e. 12-1)
hence NOT suff.

stmnt2: m^2 has 5 digits
==> \(10^4 \(10^2 \(10^6 \(m^3\) can have minimum of 7 (i.e 6+1) and max of 8 digits (i.e. 9-1)
hence NOT suff.

stmnt1&2 together: We can conclude that \(m^3\) can have minimum of 6 and max of 8 digits(i.e. 12-1) ==> m can have 7 or 8 digits
hence NOT suff.

General Discussion

If m is a positive integer, then m^3 has how many digits?
1. m has 3 digits
2. m^2 has 5 digits

How would you do this quickly? Is there a rule that I am unaware of? I could do it, but I had to pick a few numbers. Thanks.

If m is a positive integer, then \(m^3\) has how many digits?
stmnt1: \(m\) has 3 digits
==> \(10^2 \(10^6 \(m^3\) can have minimum of 6 and max of 11 digits (i.e. 12-1)
hence NOT suff.

stmnt2: m^2 has 5 digits
==> \(10^4 \(10^2 \(10^6 \(m^3\) can have minimum of 6 and max of 8 digits (i.e. 9-1)
hence NOT suff.

m^3 can have only 7 or 8 digits, not 6. If m=100=10^2 then m^3=10^6 and it has 6 trailing zeros but 7 digits.
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If m is a positive integer, then m^3 has how many digits?
1. m has 3 digits
2. m^2 has 5 digits

How would you do this quickly? Is there a rule that I am unaware of? I could do it, but I had to pick a few numbers. Thanks.

1. m has 3 digits

When I look at such statements, I invariably think of the extremities. (as muralimba did above)
Smallest m = 100 which implies m^3 = 10^6 giving 7 digits.
Largest m = 999 but it is not easy to find its cube so I take a number close to it i.e. 1000 and find its cube which is 10^9 i.e. smallest 10 digit number. Hence 999^3 will have 9 digits.
Since we can have 7, 8 or 9 digits, this statement is not sufficient.

2. m^2 has 5 digits
Now try to forget what you read above. Just focus on this statement.
Smallest m^2 = 10000 which implies m = 100
Largest m^2 is less than 99999 which gives m as something above 300 but less than 400.
Now, if m is 100, m^3 = 10^6 giving 7 digits.
If m is 300, m^3 = 27000000 giving 8 digits.
Since we have 7 or 8 digits for m, this statement is not sufficient.

e.g. statement 1 tells us that m has 3 digits. Statement 2 tells us that m is between 100 and 300 something, so statement 2 tells us that m has 3 digits (what statement 1 told us) and something extra (that its value lies between 100 and 300 something). Statement 2 is more informative and is not sufficient alone. Since statement 1 doesn’t add any new information to statement 2, no way will they both together be sufficient.
Hence answer (E).
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How do I find out how many digits in an integer?

Is there a short way to determine how many digits are in an integer? I’m new to C++, and writing a program that’s supposed to a number in one column, and its square in the next. I’m using setw to manipulate the width of the coulms.

Here is those code:

This works fine as is, but when I change MAX to 1000, the output gets screwed. I know it’s because neither column is wide enough to hold the output, so they get jammed together. I want to make the program robust, so that the output would scale to accomodate shrinking and growing MAX.
I have an idea that if I can do something like this it would work:

6 Contributors 12 Replies 446 Views 2 Years Discussion Span Latest Post 11 Years Ago Latest Post by Student level

>>Is there a short way to determine how many digits are in an integer
In a loop keep dividing it by 10 until the result is 0

> I think you misunderstood the other poster.
> I don’t think that word means what you think it means.
I amaze at the arrogance and rudeness some people possess. Do you really believe yourself superior to AD’s intellect, and mine?

I don’t really care much what you think. …

All 12 Replies

>>Is there a short way to determine how many digits are in an integer
In a loop keep dividing it by 10 until the result is 0

I thought of writing a function to do that, but I was sort of looking for an inline solution to see maybe if it was possible. I feel like I’m missing what the book is trying to get me to learn here. Anyway for the sake of moving on I’lll roll with this.

Question’s still open for anyone who might know an inline solution.

Inline solution below

Question’s still open for anyone who might know an inline solution.

Also, you want to set the width to the width of the maximum number you’ll have to print in each column. Setting it to the width of i or square is probably not what you want, because the widths of your columns will change for different values.

I think you misunderstood the other poster.

He didn’t misunderstand.

There are two ways to look at «inline»:
1. the OP wants to avoid using function call syntax
2. AD’s code actually avoids calling a function

Either way the best solution takes multiple lines of code, as posted.

I don’t think that word means what you think it means.

There are two ways to look at «inline»:

The OP was using the way AD wasn’t.

I think you misunderstood the other poster.

> I think you misunderstood the other poster.
> I don’t think that word means what you think it means.
I amaze at the arrogance and rudeness some people possess. Do you really believe yourself superior to AD’s intellect, and mine?

I don’t really care much what you think. I know what the word inline means. Look it up yourself.

> The OP was using the way AD wasn’t.
Thank you Captain Obvious.

> I think you misunderstood the other poster.
> I don’t think that word means what you think it means.
I amaze at the arrogance and rudeness some people possess.

I don’t understand why you would be touched off by straightforward responses. Also, haven’t you ever seen The Princess Bride?

Do you really believe yourself superior to AD’s intellect, and mine?

Relative to AD’s, I don’t know.

Ok ok. Just to clarify I wanted to see if I could’ve done it all within the same function, maybe inline was the wrong word to use. I was hoping someone would point out some sort of manipulator or function or something that I might have missed and was supposed to learn.

Who knows, maybe the book meant for me to use a function, come to think of it.

If anyone has the book Accelerated C++ it’s one of the exercises at the end of Chapter Four (or five if you don’t start counting at zero). Anyway thanks AD, although Sarehu is right in that the columns still get screwed because I would have to know the length of the widest one before hand. I guess I have to calculate them all first and then store them and print them back.

Thanks for the help guys. I’m marking this as solved, since the solution I had in mind looks quite unreasonable in hindsight.

EDIT: instead of calculating before hand and storing, if I base the length calculations off MAX it should work just fine 🙂

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Find out how many binary digits a particular integer has [duplicate]

What is the fastest possible way to find out how many binary digits a particular integer has when it is converted from decimal to binary in C/C++?

So 47 has 6 digits represented in binary.

9 Answers 9

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For a quick fun way of doing this without needing to call math functions, check this one out:

As a bonus, this should cook down to a memory load and 2 registers in use, for extra whiz-bang.

If you’re looking for the «fastest» way in terms of performance, you will need to resort to platform-specific methods.

Some architectures actually have an instruction that does that.

On x86, you have the bsr instruction.

In MSVC, it is accessible as:

The fastest solution that’s presented at my favorite collection of bit twiddling hacks is Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup. It requires 13 instructions to find the highest set bit in a number.

The traditional way

You can get more fancy with optimization.

EDIT 2 Here’s the fastest code I could think of without the use of Bit Scan Reverse opcode. You could use a bigger (256 entry) LUT and remove the last IF statement. In my testing this was faster than the repeated OR-SHIFT then LUT method described in another post.

Or if your in x86 or x86-64 bit architecture you can use the BSR (Bit Scan Reverse) opcode. You can find the c++ intrinsic for it http://msdn.microsoft.com/en-us/library/fbxyd7zd%28v=vs.80%29.aspx

EDIT Why the log2 answers are not optimal. While mathematically correct, complex floating point operations (sine, cosine, tan, log) are the slowest performing operations on modern computers. This is compounded by having to convert integer to a float and having to ceiling/floor it as well.