# Overview

A few short algorithms play essential roles in multiple Project Euler problems.
Instead of writing them from scratch over and over again, I copy'n'paste'n'modify where appropriate.

My goal is that all solutions include only standard C++ library headers (or C++11/14/17/you-know-the-game).
They are "single-file-only" programs without external dependencies.
Therefore the snippets shown below are NOT put into a dedicated toolbox.h file.

# Greatest Common Divisor

       // greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}


The Euclidean algorithm is a very efficient, simple (and surprisingly old !) algorithm. It's younger sibling, the binary GCD algorithm is supposed to be a tad bit faster, though.

Note: On newer versions of the GCC compiler there is a __gcd(a, b) function in #include <stl_algo.h> which might be faster on your computer.
(it's very likely that std::gcd will be added to the C++17 standard)

Modern CPU's have a few instructions that can accelerate the GCD computation by about 30%. However, accessing these instructions isn't portable.
An implementation for the GCC compiler looks like this (with automatic fallback to default algorithm on other compilers):

       // greatest common divisor
template <typename T>
T gcd(T a, T b)
{
#ifdef __GNUC__
// and          https://github.com/lemire/Code-used-on-Daniel-Lemire-s-blog/blob/master/2013/12/26/gcd.cpp
if (a == 0)
return b;
if (b == 0)
return a;

auto shift = __builtin_ctz(a | b);
a >>= __builtin_ctz(a);
do
{
b >>= __builtin_ctz(b);
if (a > b)
std::swap(a, b);

b -= a;
} while (b != 0);
return a << shift;

#else

// standard GCD
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
#endif
}


# Least Common Multiple

       // least common multiple
template <typename T>
T lcm(T a, T b)
{
// parentheses avoid overflows for certain input values
return a * (b / gcd(a, b));
}


The Least Common Multiple builds upon the Greatest Common Divisor.

# Coprime Test

       // return true if a and b are coprime
// note: needs an implementation of gcd()
template <typename T>
bool isCoprime(T a, T b)
{
// fast reject if both are even (=> gcd(a,b) >= 2)
if (((a|b) & 1) == 0)
return false;

return gcd(a, b) == 1;
}


# Prime Sieves

## Sieve Of Eratosthenes

• faster than other simple sieves
• ok-ish memory consumption
• optimized version needs a dedicated lookup-function
       #include <vector>

// 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) + 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;
}
}
}


My optimized code needs 0.5 bits per number which is less than what is required for the sieve based on trial division for most use cases.
All prime numbers from 2 to 100 million can be found in about one third of a second and need about 6 MByte.
A multi-threaded version is much faster, though.

Note: I wrote a segmented prime sieve in problem 196.

## Based On Trial Division

• simple code
• can be computed in an incremental way ("on demand")
• slow division
• not very memory-efficient (allocation is slow, too !)
       std::set<unsigned int> primes = { 2 };
for (unsigned int i = 3; i <= limit; 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.insert(i);
}


While std::set offers some nice features "for free", such as looking up whether it contains an element, its memory efficiency can be sub-prime. In those cases I replace it by std::vector and change both inserts to push_back.

Sometimes std::unordered_set might be an option, too.

## Precomputed Tables

If code size and/or compilation times are no issue, then a precomputed table of prime numbers is certainly the fastest approach:

       const unsigned int primes[] = {
#include "primes1000000.txt"
};


All files contain a comma-separated list of integers, e.g. primes100.txt looks like this:

       2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97


# Primality Tests

## Trial Division

This test is basically the same used in the Trial Division Prime Sieve, except that we don't know all smaller primes and therefore have to check against all odd numbers.

       bool isPrime(unsigned int x)
{
// even number ? only even prime is 2
if ((x & 1) == 0)
return x == 2;

// check all even numbers from 3 to sqrt(x)
for (unsigned int i = 3; i*i <= x; i += 2)
if (x % i == 0)
return false;

// passed all tests, must be a prime number
return x > 1;
}


## Wheel Factorization

Wheel Factorization reduces the number of divisions by observing that all prime factors can be expressed as:
2, 3, 5 or any m: 30m+1, 30m+7, 30m+11, 30m+13, 30m+17, 30m+19, 30m+23, 30m+29 (wheel size 30=2*3*5).
Some of these numbers are actually composite numbers (and not true prime factors) and therefore cause redundant work - but at least we don't forget about any prime factor.
More formally: this set of numbers contains all prime factors but is not minimal.

Note: running a Wheel-based primality test on all numbers from 1 to 10 million is roughly two times faster than with Trial Division
(1.8 vs. 3.3 seconds ... and only 1.4 seconds for the Miller-Rabin test).

       bool isPrime(unsigned long long x)
{
// prime test for 2, 3 and 5 and their multiples
if (x % 2 == 0 || x % 3 == 0 || x % 5 == 0)
return x == 2 || x == 3 || x == 5;

// wheel with size 30 (=2*3*5):
// test against 30m+1, 30m+7, 30m+11, 30m+13, 30m+17, 30m+19, 30m+23, 30m+29
// their deltas/increments are:
const unsigned int Delta[] = { 6, 4, 2, 4, 2, 4, 6, 2 };
unsigned long long i = 7;
// 7 belongs to the second test group
auto pos = 1;

// check numbers up to sqrt(x)
while (i*i <= x)
{
// not prime ?
if (x % i == 0)
return false;

// skip forward to next test divisor
i += Delta[pos];
// next delta/increment
pos = (pos + 1) & 7;
}

// passed all tests, must be a prime number
return x > 1;
}


## Miller-Rabin Test

The Miller-Rabin test is a probabilistic primality test. Using specific witnesses it can be made a deterministic test for all 64 bit numbers.
My code needs the mulmod and powmod function also found on this page (see here).

       // Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)

// some code from             https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from    http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/

// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 <<  2) | (1 <<  3) | (1 <<  5) | (1 <<  7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;

if (p %  2 == 0 || p %  3 == 0 || p %  5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;

if (p < 17*19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;

// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };

// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;

// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}

// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;

// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = mulmod(x, x, p);
// x % p == 1 => not prime
if (x == 1)
return false;

// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}

// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);

// prime
return true;
}


## Fermat Test

The Fermat primality test is neither fast nor accurate - but its implementation is quite short.
My code needs the mulmod and powmod function also found on this page (see here).

       // return true if p is probably prime, accuracy depends on the number of iterations
// unfortunately, so-called Carmichael numbers pass all tests, see https://en.wikipedia.org/wiki/Carmichael_number
bool isPrime(unsigned long long p, unsigned int iterations = 90)
{
// fast check against small primes, eliminates a few Carmichael numbers, too
if (p < 2 || p % 2 == 0 || p % 3 == 0 || p % 5 == 0 || p % 7 == 0 || p % 11 == 0)
return p == 2 || p == 3 || p == 5 || p == 7 || p == 11;

// just a pseudo-random seed
unsigned long long a = p / 2 + 1;

// check p against several pseudo-random numbers
for (unsigned int i = 0; i < iterations; i++)
{
// we test against random numbers between 1 and p - 1 (inclusive)
// let's compute a simple hash: (i * 2654435761) % 2^32
// (Knuth's multiplicative hash)
a *= 2654435761;
if (a == 0 || a >= p)
a = a % (p - 1) + 1;

// Fermat found that for each prime: a^(p-1) % p == 1
if (powmod(a, p-1, p) != 1)
return false; // failed test, p is definitely not prime ...
}

// p is probably prime
return true;
}


# Prime Counting

The source code for all three algorithm can be found below.

## Binary Search

If the number of primes is small (a few million) then a standard prime sieve can precompute all prime numbers.
When those primes are stored in a std::vector named primes then a basic bisection is extremely fast for an existing prime x:

       std::distance(primes.begin(), std::upper_bound(primes.begin(), primes.end(), x));

This code is actually part of the next two algorithms, too (see the first lines of the pi function).

## Legendre Algorithm

Two functions pi and phi are essential for the Legendre algorithm:
pi(x) counts all prime numbers up to (and including) x.
phi(x, a) counts all numbers up to (and including) x that are coprime to the first a primes.
The only difference between the Legendre and the faster Meissel-Lehmer algorithm is the implementation of pi.

Note: my version of phi contains the optimizations of Mapes and I added a cache for small values (phicache).
The code works perfectly fine without the neat tricks - just slower.

## Meissel-Lehmer Algorithm

The Meissel-Lehmer algorithm has a more complex pi function which puts less work on phi.

       #include <vector>
#include <algorithm>
#include <cmath>

// uncomment to run (slower) Legendre algorithm
//#define LEGENDRE

// cache small value of phi(x,a)
static std::vector<std::vector<unsigned short>> phicache;

// forward declaration
unsigned long long pi(unsigned long long x);

// return how many numbers are coprime to the first a primes
long long phi(unsigned long long x, unsigned int a, int sign = +1)
{
// Mapes equations for small a
if (a == 0)
return sign * x;
if (a == 1)
return sign * (x - x / 2);
if (a == 2)
return sign * (x - x / 2 - x / 3 + x / 6);

// precomputed tables for small x
if (x < phicache.size() && a < phicache[x].size())
return sign * phicache[x][a];

#ifndef LEGENDRE
// http://www.cnblogs.com/boceng/p/7222701.html
if (x < primes[a] * primes[a])
return sign * ((int)pi(x) - (int)a + 1);
#endif

// partially iterative algorithm
// most people use the slower version: phi(x, a-1) - phi(x/p[a], a-1)
unsigned long long result = 0;
while (a > 3) // a = 0..3 handled by Mapes equations
{
if (a < primes.size() && x < primes[a]) // check for primes.size() only needed for Legendre
return result + sign;

a--;
result += phi(x / primes[a], a, -sign);
}

// Mapes a = 3
return result + sign * (x - x / 2 - x / 3 + x / 6 - x / 5 + x / 10 + x / 15 - x / 30);
}

// two implementations:
// - Legendre: simpler but slower ==> pi(10^11) in 11 seconds
// - Lehmer:   more complex but faster => pi(10^11) in about 0.7 seconds
// both numbers including the overhead of the prime sieve (up to sqrt(10^11)) and caching pi up to 100 and phi up to (1000,168)
unsigned long long pi(unsigned long long x)
{
// small x: fast binary search in list of primes
if (x < primes.back())
{
auto nextPrime = std::upper_bound(primes.begin(), primes.end(), x);
return std::distance(primes.begin(), nextPrime);
}

#ifdef  LEGENDRE
// Legendre
auto a = pi(sqrt(x));
return phi(x, a) + a - 1;
#else
// Meissel-Lehmer
// good overview: http://acganesh.com/blog/2016/12/23/prime-counting

auto a = pi(pow(x, 0.25));
auto b = pi(sqrt(x));
auto c = pi(cbrt(x));

auto result = phi(x, a) + (b + a - 2) * (b - a + 1) / 2;
for (auto i = a; i < b; i++)
{
auto xDivP = x / primes[i];
result -= pi(xDivP);

if (i < c)
{
auto b_i = pi(sqrt(xDivP));
for (auto j = i; j < b_i; j++)
result -= pi(xDivP / primes[j]) - j;
}
}
return result;
#endif
}


## Benchmark

The function countPrimes in problem 501 does the same job - and is much faster.
However, I didn't write it (only modified it a little bit) and still have no complete understanding of its inner workings despite Muthu Veerappan's explanation on this blog.

pi(1011) = 4118054813
Legendre 11.2 seconds
Meissel-Lehmer 0.7 seconds
PE501 0.25 seconds

pi(1012) = 37607912018
Legendre 103 seconds,
Meissel-Lehmer 4.3 seconds
PE501 1.2 seconds

# Modular Arithmetic

## Modular Multiplication

There are multiple versions of my modular arithmetic multiplication code (mulmod) which solves (a*b)%modulocode>.
The 32 bit version is extremely short:

       // return (a*b) % modulo
unsigned int mulmod(unsigned int a, unsigned int b, unsigned int modulo)
{
return ((unsigned long long)a * b) % modulo;
}


The story is completely different for 64 bit because of a lack of standardized 128 bit integers in C/C++.
Bitwise multiplication is very similar to multiplication taught in school (and works only up to 63 bits):

       // return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;

// next bit
b >>= 1;
}

return result;
}


The blockwise algorithm is an extension of the bitwise algorithm and multiplies multiple bits at once.
The 63-bit limit applies as well:

       // return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// count leading zero bits of modulo
unsigned long long m = modulo;
while ((m & 0x8000000000000000ULL) == 0)
{
m <<= 1;
}

// cover all bits of modulo

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
result += (b & mask) * a;
result %= modulo;

// next bits
a  %= modulo;
}
return result;
}


A GCC-only version of mulmod can be considerably faster:

       unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// note: you can remove the following two lines
// but if (a*b)/c > 2^64 then the Assembler code will trigger a floating-point exception
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// perform 128 bit arithmetic

#ifdef __x86_64__
// use GCC inline assembler
unsigned long long asmResult;
__asm__
(
"mulq %2\n" // result in rdx:rax
"divq %3"   // quotient in rax, remainder in rdx
: "=&d" (asmResult), "+%a" (a)
: "rm" (b), "rm" (modulo)
: "cc"      // clear conditions
);
return asmResult;

#else

// based on GCC's 128 bit implementation
return ((unsigned __int128)a * b) % modulo;
#endif
}


## Modular Exponentiation

powmod solves (base^exponent)%modulo. My algorithm is based on Exponentiation by squaring.
The 32-bit version doesn't have any dependencies:

       // return (base^exponent) % modulo for 32-bit values, no need for mulmod
unsigned int powmod(unsigned int base, unsigned int exponent, unsigned int modulo)
{
unsigned int result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = (result * (unsigned long long)base) % modulo;

// even exponent ? a^b = (a*a)^(b/2)
base = (base * (unsigned long long)base) % modulo;
exponent >>= 1;
}
return result;
}


The 64-bit version needs mulmod (see above):

       // return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}


## Modular Inverse

The Extended Euclidean algorithm is the fastest way and comes with the least restrictions. It's code looks a bit messy, though.
If the modulo is a prime number then thanks to Fermat's little theorem I can find the modular inverse with powmod:

       // extended Euclidean algorithm, see https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm
unsigned int modularInverse(unsigned int a, unsigned int modulo)
{
// pretty much the same code can be found on https://rosettacode.org/wiki/Modular_inverse
auto originalModulo = modulo;
// note: s and t can be negative inside the loop
int s = 0;
int t = 1;
while (a > 1)
{
auto tmp      =     modulo;
auto quotient = a / modulo;
modulo        = a % modulo;
a = tmp;

auto tmp2 = s;
s = t - quotient * s;
t = tmp2;
}

// avoid negative result
return t < 0 ? t + originalModulo : t;
}

// Fermat's little theorem, see https://en.wikipedia.org/wiki/Fermat%27s_little_theorem
// modulo needs to be a prime number
unsigned int modularInverseOfPrime(unsigned int a, unsigned int modulo)
{
// Fermat's Little Theorem: a^p     == a mod p
// divide both sides by a:  a^(p-1) == 1 mod p
// once more:               a^(p-2) == a^-1 mod p
// where a^-1 is the inverse of a => equal to a^(p-2) % p
return powmod(a, modulo - 2, modulo);
}


# Pseudo-Random Numbers

For most of the statistics / probability problems I write a simple Monte-Carlo simulation in order to estimate the result (to avoid that my "proper" algorithm doesn't go wild).
The implementation of the rand() function from the C standard library varies among compilers / platforms.
Therefore I prefer this simple code to reproduce exactly the same result on different machines:

## Linear Congruential Generator

       // a simple pseudo-random number generator
// (produces the same result no matter what compiler you have - unlike rand() from math.h)
unsigned int myrand()
{
static unsigned long long seed = 0;
seed = 6364136223846793005ULL * seed + 1;
return (unsigned int)(seed >> 30);
}


The output isn't on a cryptographically safe level but more than sufficient for the Project Euler problems. And pretty fast !

# Working With Digits

## Digital Sum

Efficient routines to analyze the digits of a number are part of many problems.
The most frequent is the digital sum (sum of all digits):

       // sum of all digits
unsigned int digitSum(unsigned long long x)
{
unsigned int result = 0;
while (x > 0)
{
result += x % 10;
x      /= 10;
}
return result;
}

// slightly more efficient (saves one iteration)
unsigned int digitSumFaster(unsigned long long x)
{
unsigned int result = 0;
while (x >= 10)
{
result += x % 10;
x      /= 10;
}
return result + x;
}


The second version is slightly more efficient but in most cases you won't notice a significant difference.
I tried std::lldiv (from the C++11 standard) instead of modulo/division but it's much slower with my G++ compiler.

## Fingerprint

In order to find out whether two numbers are permutations of each other, I generate their "fingerprint":
I count how often each digit can be found and return a number where the lowest 5 bits represent the amount of zeros,
the next 5 bits the amount of ones, then twos, etc.
Note: the term "fingerprint" isn't something official - I just like the word.

       // count digits, two numbers share the same fingerprint if they are permutations of each other
unsigned long long fingerprint(unsigned long long x)
{
unsigned long long result = 0;
while (x > 0)
{
auto digit = x % 10;
x /= 10;

result += 1ULL << (5 * digit);
}
return result;
}


# Combinatorics

## Binomial Coefficient

The formula for a binomial coefficient is pretty simple but easily causes overflows due to the factorial function.
The following code can compute most of the values if the result fits in 64 bits (even though the factorials may exceed 64 bits):

       // number of ways to choose n elements from k available
unsigned long long choose(unsigned int n, unsigned int k)
{
// n! / (n-k)!k!
unsigned long long result = 1;
// choose(n, 0) = 1
// choose(n, k) = choose(n, k - 1) * (n + 1 - k) / k
// reduce overflow by dividing as soon as possible to keep numbers small
for (unsigned int invK = 1; invK <= k; invK++)
{
result *= n;
result /= invK;
n--;
}
return result;
}


# Big Integers

Working with very large numbers can be hard in C++ (whereas it's extremely simple in Java and Python).
Of course, there are dedicated C/C++ libraries, such as GMP, but I don't want to link/include external libraries.

Addition, subtraction and multiplication are straightforward implementations of basic school algorithms.
MaxDigit can be set to other powers-of-10 in order to speed up the code but then the std::string functions will fail.




# Fractions

The C++ STL class std::ratio supports only compile-time arithmetic.
Therefore I had to write my class Fraction which was tuned for performance:

• it doesn't automatically reduce fractions (to become proper fractions), you have to explicitly call reduced()
• there are no checks whether the denominator is zero nor are the signs of numerator / denominator normalized
• overflows do not cause exceptions and will produce undefined results
• adjust the data type T to fit your needs, e.g. reduced() is much slower for long long then int

       // simple class to representation a fraction
// note: no checks for a zero denominator
//       signs are not normalized
struct Fraction
{
// change to long long if you need to support larger values
typedef int T;

// numerator
T num;
// denominator
T den;

Fraction(T numerator, T denominator = 1)
: num(numerator), den(denominator)
{}

Fraction operator+(const Fraction& other) const
{
if (den == other.den)
return Fraction(num + other.num, den);
// n1/d1 + n2/d2 = (n1*d2 + n2*d1) / d1*d2
return { num * other.den + other.num * den, den * other.den };
}
// subtract
Fraction operator-(const Fraction& other) const
{
if (den == other.den)
return Fraction(num - other.num, den);
// n1/d1 - n2/d2 = (n1*d2 - n2*d1) / d1*d2
return { num * other.den - other.num * den, den * other.den };
}
// multiply
Fraction operator*(const Fraction& other) const
{
// n1/d1 * n2/d2 = n1*n2 / d1*d2
return Fraction(num * other.num, den * other.den);
}
// divide
Fraction operator/(const Fraction& other) const
{
// n1/d1 / n2/d2 = n1*d2 / d1*n2
return Fraction(num * other.den, den * other.num);
}
// sort
bool operator< (const Fraction& other) const
{
// n1/d1 < n2/d2 => n1*d2 < n2*d2
return num * other.den <  other.num * den;
}
// compare
bool operator==(const Fraction& other) const
{
// n1/d1 < n2/d2 => n1*d2 < n2*d2
return num * other.den == other.num * den;
}

// return Fraction with swapped numerator and denominator
Fraction inverse() const
{
return { den, num };
}

// convert to a proper fraction
Fraction reduced() const
{
// catch simple cases: 0/d and n/1
if (num == 0 || den == 1)
return { num, 1 };

// divide numerator and denominator by their gcd()
// code taken from my toolbox / gcd()
auto a = num;
auto b = den;
while (a != 0)
{
auto c = a;
a = b % a;
b = c;
}
// now b contains the gcd()
if (b == 1)
return *this;
return { num/b, den/b };
}
};


# 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 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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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !