<< problem 86 - Cuboid route Product-sum numbers - problem 88 >>

# Problem 87: Prime power triples

The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28.
In fact, there are exactly four numbers below fifty that can be expressed in such a way:

28 = 2^2 + 2^3 + 2^4
33 = 3^2 + 2^3 + 2^4
49 = 5^2 + 2^3 + 2^4
47 = 2^2 + 3^3 + 2^4

How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?

# Algorithm

A simple prime sieve is responsible to find all primes i < sqrt{50000000}.
Then three nested loops compute all sums of all combinations of such primes where a^2 + b^3 + c^4 < 50000000.
Be careful: b*b*b and c*c*c*c can easily exceed an unsigned int.

Those sums are sorted and duplicate sums are removed. Now we have a nice std::vector with all sums.
The original problem is solved now (just display sums.size()) but the Hackerrank problem is slightly tougher.

## Modifications by HackerRank

In order to find the the number of sums which are below a certain input value, I search through the sorted container with std::upper_bound
and then compute the distance to the beginning of the container.
That's extremely fast (std::upper_bound most likely uses binary search) and easily processed thousands of test cases per second.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <vector>
#include <algorithm>
#include <iostream>

int main()
{
const unsigned int MaxLimit = 100 * 1000 * 1000; // Hackerrank: 10^7 instead of 5*10^6

// prime sieve
std::vector<unsigned int> primes;
primes.push_back(2);
for (unsigned int i = 3; i*i < MaxLimit; i += 2)
{
bool isPrime = true;

// test against all prime numbers we have so far (in ascending order)
for (auto p : primes)
{
// next prime is too large to be a divisor ?
if (p*p > i)
break;

// divisible ? => not prime
if (i % p == 0)
{
isPrime = false;
break;
}
}

// yes, we have a prime number
if (isPrime)
primes.push_back(i);
}

// just three nested loops where I generate all sums
std::vector<unsigned int> sums;
for (auto a : primes)
for (auto b : primes)
for (auto c : primes)
{
auto a2 = a*a;
auto b3 = (unsigned long long)b*b*b;
auto c4 = (unsigned long long)c*c*c*c;
auto sum = a2 + b3 + c4;
// abort if too big
if (sum > MaxLimit)
break;

sums.push_back(sum);
}

// sort ascendingly
std::sort(sums.begin(), sums.end());
// a few sums occur twice, let's remove them !
auto last = std::unique(sums.begin(), sums.end());

// process test cases
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int limit = MaxLimit;
std::cin >> limit;

// find next sum which is bigger than the limit
auto pos = std::upper_bound(sums.begin(), last, limit);
// how many sums are inbetween 28 and limit ?
auto num = std::distance(sums.begin(), pos);
std::cout << num << std::endl;
}

return 0;
}


This solution contains 11 empty lines, 12 comments and 3 preprocessor commands.

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 50" | ./87

Output:

Note: the original problem's input 50000000 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 0.16 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 20 MByte.

(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

March 12, 2017 submitted solution

# Hackerrank

My code solves 9 out of 9 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 20% (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 never an option.

projecteuler.net/thread=87 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-87-sum-power/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p087.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler087.scala (written by Michael Bayne)

# 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.

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

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The 160 solved problems had an average difficulty of 21.8% at Project Euler and I scored 11,807 points (out of 13100) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
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