<< problem 232 - The Race | An Arithmetic Geometric sequence - problem 235 >> |
Problem 234: Semidivisible numbers
(see projecteuler.net/problem=234)
For an integer n >= 4, we define the lower prime square root of n, denoted by lps(n), as the largest prime <= sqrt{n}
and the upper prime square root of n, ups(n), as the smallest prime >= sqrt{n}.
So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37.
Let us call an integer n >= 4 semidivisible, if one of lps(n) and ups(n) divides n, but not both.
The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8, 10 and 12.
15 is not semidivisible because it is a multiple of both lps(15) = 3 and ups(15) = 5.
As a further example, the sum of the 92 semidivisible numbers up to 1000 is 34825.
What is the sum of all semidivisible numbers not exceeding 999966663333 ?
My Algorithm
My standard prime sieve from my toolbox is extremely quick in finding all prime numbers below sqrt{999966663333} approx 10^6.
I always look at two consecutive primes numbers last
and next
.
Numbers between last^2 and next^2 are semidivisible if they are either a multiple of last or next but not both.
Similar to the very first problem at Project Euler (problem 1) I count all multiples of last and next and subtract all multiples of last * next.
I tried to find a closed form but they all looked rather messy. The simple for
-loops perform the same task in less than 0.02 seconds.
The hardest part was to get the borders right, especially the last interval from 999983^2 = 999966000289 to 999966663333.
That's why there are (too many ?) checks against limit
.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 1000 | ./234
Output:
Note: the original problem's input 999966663333
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)
My code
… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.
#include <iostream>
#include <vector>
#include <cmath>
// ---------- standard prime sieve from my toolbox ----------
// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;
// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;
// lookup for odd numbers
return sieve[x >> 1];
}
// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = size >> 1;
// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;
// process all relevant prime factors
for (unsigned int i = 1; 2*i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3*i+1;
while (current < half)
{
sieve[current] = false;
current += 2*i+1;
}
}
}
int main()
{
unsigned long long limit = 999966663333ULL;
std::cin >> limit;
fillSieve(sqrt(limit) + 100); // a few more primes than strictly necessary ...
// sum of all matches
unsigned long long sum = 0;
// 2 is the smallest prime
unsigned long long last = 2;
// until lps(last^2) exceeds the limit
while (last*last <= limit)
{
// find next prime
auto next = last + 1;
while (!isPrime(next))
next++;
// note: for each square of a prime lps(p^2) = ups(p^2) = p
// => p^2 is obviously divisible by both and hence can't be semidivisible
// all loops run from "from" and "to" but skip both (because not semidivisible)
auto from = last * last;
auto to = next * next;
// the following loops could be replaced by clever arithmetic
// but they are super-fast and most of the time is spent sieving primes anyway
// divisible by lps(i)
for (auto i = from + last; i < to && i <= limit; i += last)
sum += i;
// don't exceed the limit
while (to - next > limit)
to -= next;
// divisible by ups(i)
for (auto i = to - next; i > from; i -= next)
sum += i;
// but not those divisible by both
for (auto i = from - from % (last * next); i < to && i <= limit; i += last * next)
{
if (i <= from)
continue;
if (i > limit)
break;
// counted twice, remove both
sum -= i + i;
}
// next pair of primes
last = next;
}
// display result
std::cout << sum << std::endl;
return 0;
}
This solution contains 19 empty lines, 28 comments and 3 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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
August 5, 2017 submitted solution
August 5, 2017 added comments
Difficulty
Project Euler ranks this problem at 50% (out of 100%).
Links
projecteuler.net/thread=234 - the best forum on the subject (note: you have to submit the correct solution first)
Heatmap
Please click on a problem's number to open my solution to that problem:
green | solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too | |
yellow | solutions score less than 100% at Hackerrank (but still solve the original problem easily) | |
gray | problems are already solved but I haven't published my solution yet | |
blue | solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much | |
orange | problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte | |
red | problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too | |
black | problems are solved but access to the solution is blocked for a few days until a new problem is published | |
the flashing problem is the one I solved most recently |
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I scored 13,486 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
Look at my progress and performance pages to get more details.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 232 - The Race | An Arithmetic Geometric sequence - problem 235 >> |