Problem 6: How Many Digits?

https://open.kattis.com/problems/howmanydigits

The number of digits of an integer can be calculated by the following formula:

\text{# of digits in }x = \begin{cases} \log_{10} (x)+1 & : x \text{ is a power of }10 \\ \lceil \log_{10} (x) \rceil & : \text{ otherwise} \end{cases}

$\lceil x \rceil$is the ceiling function, which rounds a number up to the closest larger integer.

Another important property of logs: $\log(xy) = \log(x) + \log(y)$.

$n!$is only a power of 10 if $n=0$ or $n=1$, so handle those cases separately. Otherwise:

$$\lceil \log_{10}n! \rceil = \lceil \log_{10} [(n)(n-1)\dots(1)] \rceil = \lceil \log_{10}(n)+\dots+\log_{10}(1)\rceil$$

Store the sums $\log_{10}(1) + \dots + \log{10}(k)$in a vector, and output the ceiling of that sum for each test case.

C++ has functions ceil(x) and log10(x) that can be used for calculations. Be careful that ceil() returns a double, so cast to int or long long before printing.