<< problem 126 - Cuboid layers | Hexagonal tile differences - problem 128 >> |
Problem 127: abc-hits
(see projecteuler.net/problem=127)
The radical of n, rad(n), is the product of distinct prime factors of n.
For example, 504 = 23 * 32 * 7, so rad(504) = 2 * 3 * 7 = 42.
We shall define the triplet of positive integers (a, b, c) to be an abc-hit if:
1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
2. a < b
3. a + b = c
4. rad(abc) < c
For example, (5, 27, 32) is an abc-hit, because:
1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
2. 5 < 27
3. 5 + 27 = 32
4. rad(4320) = 30 < 32
It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for c < 1000, with sum{c} = 12523.
Find sum{c} for c < 120000.
My Algorithm
My first step is to produce all rad[]
in a sieve-like manner.
Then two nested loops process each pair (a,b) where a < b and a + b < 120000.
Check whether rad[a*b*c] < c
:
The problem statement clearly states that gcd(a,b)=gcd(b,c)=gcd(a,c)=1. If a, b and c have no shared divisors, then their rad have no shared divisors, too.
It follows that rad[a*b*c] = rad[a] * rad[b] * rad[c]
.
If gcd(a,b)=1 then gcd(a,c)=gcd(a,b+a)=1 and gcd(b,c)=gcd(c,b)=gcd(a+b,b)=1. It's sufficient to test only gcd(a,b) == 1
.
A minor speedup comes from the observation that a and b can't be even at the same time because then they would both contain prime factor 2:
if a is even then incrementB
ensures that b is always odd.
A second trick is that gcd(a, b) != 1 only if a and b share at least one prime factor. However, all prime factors can be found in rad(a) and rad(b), too.
Therefore if gcd(rad(a), rad(b)) = 1 then gcd(a, b) = 1. That's helpful since the iterative gcd()
algorithm is faster with smaller parameters and rad(x) <= x.
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 | ./127
Output:
Note: the original problem's input 120000
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>
// greatest common divisor
unsigned int gcd(unsigned int a, unsigned int b)
{
while (a != 0)
{
unsigned int c = a;
a = b % a;
b = c;
}
return b;
}
int main()
{
unsigned int limit = 120000;
std::cin >> limit;
// sieve-like computation of rad
std::vector<unsigned int> rad(limit, 1);
for (unsigned int i = 2; i < limit; i++)
{
// if rad(i) > 1 then "a" is not prime
if (rad[i] > 1)
continue;
// process all multiples of a
for (unsigned int j = 1; i*j < limit; j++)
rad[i*j] *= i;
}
// walk through all potential a and b
unsigned long long sum = 0;
for (unsigned int a = 1; a < limit/2; a++)
{
unsigned int incrementB = 1;
// make sure that if a is even then b is always odd
if (a % 2 == 0)
incrementB = 2;
for (unsigned int b = a + 1; a + b < limit; b += incrementB)
{
auto c = a + b;
// this check is much faster than the gcd()-computation and eliminates most candidates
auto prodRad = rad[a] * rad[b] * (unsigned long long)rad[c];
if (prodRad >= c)
continue;
// no shared prime factors
if (gcd(rad[a], rad[b]) != 1)
continue;
// yes, found another abc-hit
sum += c;
}
}
// display result
std::cout << sum << std::endl;
return 0;
}
This solution contains 11 empty lines, 10 comments and 3 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 2.4 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
June 26, 2017 submitted solution
June 26, 2017 added comments
Difficulty
Project Euler ranks this problem at 50% (out of 100%).
Links
projecteuler.net/thread=127 - 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:
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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 | |
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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 126 - Cuboid layers | Hexagonal tile differences - problem 128 >> |