@admin
can you please make my task public before 2054 ? thx in advance
can you please make my task public before 2054 ? thx in advance
my task
hehe
what is the name of the task?
Oh, i see, it is the number of digits.
I made it public. However, I must agree that the task i really, really easy.
My solution has three lines of code.
Sorry, I forgot that this wasn't the moderator forum :(
I made it public. However, I must agree that the task i really, really easy.
My solution has three lines of code.
Sorry, I forgot that this wasn't the moderator forum :(
OMFG
It's the one of the task that we wanted to add on our competition :(
It's the one of the task that we wanted to add on our competition :(
@duxassassin
sorry to disappoint you, i knew this formula long time ago, but i didn't know how to express decimal log with ln until 2 months ago... then i decided to upload it :)
sorry to disappoint you, i knew this formula long time ago, but i didn't know how to express decimal log with ln until 2 months ago... then i decided to upload it :)
admin owned Nikola94 xD
i knew how to calculate the number, by using log(10), but i didn't know that lg a = ln a / ln 10...
i have to say that i hate tasks like this one ...
so there aren't any idea to solve it ...
it's just question: do you know formula ? ...
and that is stupid for me, but that is just my opinion ...
so there aren't any idea to solve it ...
it's just question: do you know formula ? ...
and that is stupid for me, but that is just my opinion ...
Sorry, I forgot that this wasn't the moderator forum :(
I agree that this task is more like a Math problem than algorithm problem.
Sorry again for posting the solution, I thought this topic was in the moderator sub-forum (not visible to everyone).
Anyway, there are different ways to solve this problem, so its not that bad for algorithm problem as well.
Sorry again for posting the solution, I thought this topic was in the moderator sub-forum (not visible to everyone).
Anyway, there are different ways to solve this problem, so its not that bad for algorithm problem as well.
can you say how to solve this task in time without that formula ???
i know that formula, but can someone explain why that formula works ???
i know that formula, but can someone explain why that formula works ???
Here is why:
You want to solve the following inequality:
10^{k} \le a^b < 10^{k+1}
The easy way to understand this is to look at the equality
\ln(x^y) = y \ln(x)
Now back to the original inequality. Since logarithm is a monotonic function, always increases. And since a^b is always bigger than zero, we can write:
\ln(10^{k})} \le \ln(a^b) < \ln(10^{k+1})
k \ln(10)} \le b \ln(a) < (k+1) \ln(10)
So we get that
k \le \frac {b \ln(a)}{\ln(10)} < (k+1)
So we get that
k = \bigg \lfloor \frac {b \ln(a)}{\ln(10)} \bigg \rfloor
And the number of digits is k+1
You want to solve the following inequality:
10^{k} \le a^b < 10^{k+1}
The easy way to understand this is to look at the equality
\ln(x^y) = y \ln(x)
Now back to the original inequality. Since logarithm is a monotonic function, always increases. And since a^b is always bigger than zero, we can write:
\ln(10^{k})} \le \ln(a^b) < \ln(10^{k+1})
k \ln(10)} \le b \ln(a) < (k+1) \ln(10)
So we get that
k \le \frac {b \ln(a)}{\ln(10)} < (k+1)
So we get that
k = \bigg \lfloor \frac {b \ln(a)}{\ln(10)} \bigg \rfloor
And the number of digits is k+1
ok, i understand ...
or easier to explain ...
a = 10^log10(a)
so a^b = (10^log10(a))^b = 10^(log10(a)*b)
so it's int(log10(a)*b)+1 ...
and if you know log10 N = log2(N)/log2(10) ...
do you know to solve that task if you don't use that formula ???
you said: "Anyway, there are different ways to solve this problem, so its not that bad for algorithm problem as well."
or easier to explain ...
a = 10^log10(a)
so a^b = (10^log10(a))^b = 10^(log10(a)*b)
so it's int(log10(a)*b)+1 ...
and if you know log10 N = log2(N)/log2(10) ...
do you know to solve that task if you don't use that formula ???
you said: "Anyway, there are different ways to solve this problem, so its not that bad for algorithm problem as well."
Here is one example:
write A as x * 10^y. with 0.1>=x>1. Now you want to find
x^b * 10^(by)
and think how can you calculate x^b accurately enough!
write A as x * 10^y. with 0.1>=x>1. Now you want to find
x^b * 10^(by)
and think how can you calculate x^b accurately enough!