<< problem 181 - Investigating in how many ways objects of two ... Number Mind - problem 185 >>

# Problem 183: Maximum product of parts

Let N be a positive integer and let N be split into k equal parts, r = N/k, so that N = r + r + ... + r.
Let P be the product of these parts, P = r * r * ... * r = r^k.

For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 + 2.2 + 2.2, then P = 2.2^5 = 51.53632.

Let M(N) = P_max for a given value of N.

It turns out that the maximum for N = 11 is found by splitting eleven into four equal parts which leads to P_max = (11/4)^4;
that is, M(11) = 14641/256 = 57.19140625, which is a terminating decimal.

However, for N = 8 the maximum is achieved by splitting it into three equal parts, so M(8) = 512/27, which is a non-terminating decimal.

Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is a terminating decimal.

For example, sum{D(N)} for 5 <= N <= 100 is 2438.

Find sum{D(N)} for 5 <= N <= 10000.

# My Algorithm

My first approach was to have a simple loop running from k = 1 to n:
- find the largest (n/k)^k
- figure out whether (n/k)^k terminates

When I looked at the values of k I saw that n/k was always approx 2.7.
And Wolfram Alpha confirmed my suspicion: www.wolframalpha.com/input/?i=max((8%2Fn)%5En)
M(8) = max((8/n)^n) = e^{8/e} at k = 8/e where e = 2.7182818284..., see en.wikipedia.org/wiki/E_(mathematical_constant)

In general: max((x/n)^n) = e^{n/e} at k = n/e
Unfortunately, n/e isn't an integer and therefore the best k is k = round(n / 2.718281828) (that's enough digits of e for this problem)

The number (n/k)^k terminates if n/k terminates.
And n/k terminates if the greatest common divisor gcd(n, k) has only 2 and 5 as prime factors (something I learnt in problem 26).

## Alternative Approaches

As mentioned before, my initial approach computed all (n/k)^k.
But numbers became huge very fast ... well, I didn't need their "true" value: it's sufficient to have their logarithm because if a < b then log(a) < log(b), too.
In the end I converted (x/n)^n to n log(x/n).

Using round can be dangerous if not enough digits of e are used. 10 digits are totally fine for that problem, though.

## Modifications by HackerRank

I need to cache intermediate result to be able to handle the huge number of input values.

# My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <vector>
#include <cmath>

int main()
{
// five dummy entries
std::vector<long long> cache(5, 0);

#define ORIGINAL
#ifndef ORIGINAL
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
#endif
{
unsigned int limit = 10000;
std::cin >> limit;

auto result = cache.back();
for (unsigned int n = cache.size(); n <= limit; n++)
{
unsigned int k = round(n / 2.718281828);

// terminating only if n/k has only prime factors 2 and 5
while (k % 5 == 0)
k /= 5;
while (k % 2 == 0)
k /= 2;

if (n % k == 0)
result -= n;
else
result += n;

// cache intermediate results, too (needed for Hackerrank only)
cache.push_back(result);
}

// display result
std::cout << cache[limit] << std::endl;
}

return 0;
}


This solution contains 8 empty lines, 4 comments and 6 preprocessor commands.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):
Note: Enter the maximum N

This is equivalent to
echo 100 | ./183

Output:

Note: the original problem's input 10000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

June 13, 2017 submitted solution
July 28, 2017 modified to solve Hackerrank, too

# Hackerrank

My code solves 11 out of 11 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 45% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

Please click on a problem's number to open my solution to that problem:

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The 239 solved problems (level 9) had an average difficulty of 29.1% at Project Euler and
I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

 << problem 181 - Investigating in how many ways objects of two ... Number Mind - problem 185 >>
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