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Problem 401: Sum of squares of divisors
(see projecteuler.net/problem=401)
The divisors of 6 are 1,2,3 and 6.
The sum of the squares of these numbers is 1+4+9+36=50.
Let sigma2(n) represent the sum of the squares of the divisors of n. Thus sigma2(6)=50.
Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=sum{sigma2(i)} for i=1 to n.
The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.
Find SIGMA2(10^15) mod 10^9.
My Algorithm
Any number i has \lfloor frac{n}{i} \rfloor multiples (integer division).
When calculating SIGMA2(6):
- 1 has \lfloor frac{6}{1} \rfloor = 6 multiples, add 6 * 1^2 = 6 * 1 = 6
- 2 has \lfloor frac{6}{2} \rfloor = 3 multiples, add 3 * 2^2 = 3 * 4 = 24
- 3 has \lfloor frac{6}{3} \rfloor = 2 multiples, add 2 * 3^2 = 2 * 9 = 18
- 4 has \lfloor frac{6}{4} \rfloor = 1 multiples, add 1 * 4^2 = 1 * 16 = 16
- 5 has \lfloor frac{6}{5} \rfloor = 1 multiples, add 1 * 5^2 = 1 * 25 = 25
- 6 has \lfloor frac{6}{6} \rfloor = 1 multiples, add 1 * 6^2 = 1 * 36 = 36
Unfortunately this requires n iterations, which is no problem for n = 6 but next to impossible for n = 15.
Looking closer at the example reveals that half of the divisors appears only once (4, 5 and 6).
In general, a few number appear very often and the majority just a few times.
Because i appears \lfloor frac{n}{i} \rfloor times, i <= \lfloor frac{n}{i} \rfloor if i <= sqrt{n}.
If i > sqrt{n} then several squares appear the same time. As mentioned before, there are three squares that appear once.
My program applies the basic algorithm for all numbers up to sqrt{n}, which needs sqrt{10^15} approx 3.2 * 10^7 iterations.
Then it determines which squares appear once, twice, etc. going backwards from 10^15 to sqrt{n}.
This can be quite efficient because the sum of a series of squares is (see en.wikipedia.org/wiki/Square_pyramidal_number):
P(n) = sum_{k=1...n}{k^2} = dfrac{n (n + 1) (2n + 1)}{6}
The largest square pyramidal number that appears exactly m times is \lfloor frac{n}{m} \rfloor.
To obtain the sum of all square that appear m times I calculate P(\lfloor frac{n}{m} \rfloor) - P(\lfloor frac{n}{m+1} \rfloor).
pyramidal(n)
returns the n-th square pyramidal number (mod 10^9). This function is rather tricky because I hit the limits of C++'s 64 bit integers.Luckily, G++ has routines for 128 bit integers to work around this issue - but therefore this code doesn't compile with Visual C++.
The code is scattered with mod statement. Maybe a few are redundant.
Note
Apparently you can implement pyramidal
without __int128
.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 6 | ./401
Output:
Note: the original problem's input 1000000000000000
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++. You can download it, too. Or just jump to my GitHub repository.
#include <iostream>
#include <cmath>
// return sum of 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 mod modulo
// note: needs GCC's 128 bit integers (not available on Visual C++)
unsigned int pyramidal(unsigned long long n, unsigned int modulo)
{
// inspired by https://stackoverflow.com/questions/39826483/square-pyramidal-number-of-int32-modulo-some-m-using-only-int64-intermediates
unsigned __int128 nn = n;
// solve n * (n + 1) * (2n + 1) / 6
// either nn or nn+1 is even => their product must be divisible by 2
auto x = nn * (nn + 1) / 2;
auto y = 2 * n + 1;
// now equation becomes x * y / 3 (mod modulo)
// is x a multiple of 3 ?
auto div3 = x / 3;
if (div3 * 3 == x)
return ((div3 % modulo) * y) % modulo;
// no, y must be a multiple of 3
return (x % modulo) * (y / 3) % modulo;
}
// count all divisors
unsigned int solve(unsigned long long n, unsigned int modulo)
{
unsigned long long sum = 0;
// for each number i there is a second divisor n/i (integer division)
// for i < sqrt(n) those second divisors are "unique"
// for i > sqrt(n) several i share the same second divisor n/i
unsigned int threshold = sqrt(n);
// look at small divisors
for (unsigned long long i = 1; i <= threshold; i++)
{
// each divisor-th number is a multiple of divisor
auto count = (n / i) % modulo;
// square (reduce with modulo)
auto square = (i * i) % modulo;
// add
sum += (count * square) % modulo;
}
// apply the same logic in reverse order:
// for each divisor i there is a divisor j = n / i
for (unsigned int j = 1; j <= n / (threshold + 1); j++)
{
auto from = n / j;
auto to = n / (j + 1);
unsigned long long sumOfConsecutiveSquares = 0;
// note: from > to, therefore decrement instead of increment
/*for (unsigned long long k = from; k > to; k--)
{
auto square = (k * k) % modulo;
sumOfConsecutiveSquares += square;
sumOfConsecutiveSquares %= modulo;
}*/
// same as the for-loop above
auto squareSumFrom = pyramidal(from, modulo); // sum of squares <= from
auto squareSumTo = pyramidal(to, modulo); // sum of squares <= to
// add modulo to prevent negative numbers
if (squareSumFrom < squareSumTo)
squareSumFrom += modulo;
sumOfConsecutiveSquares = squareSumFrom - squareSumTo;
// those squares can be found "j" times
sum += (j * sumOfConsecutiveSquares) % modulo;
}
// and again, take care of modulo
return sum % modulo;
}
int main()
{
// 10^9
const auto Modulo = 1000000000;
// 10^15
auto n = 1000000000000000ULL;
std::cin >> n;
std::cout << solve(n, Modulo) << std::endl;
return 0;
}
This solution contains 13 empty lines, 25 comments and 2 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 3.8 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 22, 2017 submitted solution
August 22, 2017 added comments
Difficulty
Project Euler ranks this problem at 25% (out of 100%).
Links
projecteuler.net/thread=401 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Python github.com/evilmucedin/project-euler/blob/master/euler401/401.py (written by Den Raskovalov)
Python github.com/Meng-Gen/ProjectEuler/blob/master/401.py (written by Meng-Gen Tsai)
Python github.com/nayuki/Project-Euler-solutions/blob/master/python/p401.py (written by Nayuki)
Python github.com/smacke/project-euler/blob/master/python/401.py (written by Stephen Macke)
C++ github.com/roosephu/project-euler/blob/master/401.cpp (written by Yuping Luo)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p401.java (written by Nayuki)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem401.java (written by Magnus Solheim Thrap)
Mathematica github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p401.mathematica (written by Nayuki)
Haskell github.com/roosephu/project-euler/blob/master/401.hs (written by Yuping Luo)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 | |
black | problems are solved but access to the solution is blocked for a few days until the next problem is published | |
[new] | the flashing problem is the one I solved most recently |
I stopped working on Project Euler problems around the time they released 617.
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I scored 13526 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 !!!
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