Add some explanation about mu function, add better gcd

Author: mehranagh20

handy-algorithms/mu.explain

@@ -2,3 +2,5 @@ mu[i] = {0 if i contains a prime more than once
  1 if i consists of even number of primes
  -1 otherwise, ie. odd number of primes}
 mu can be used to count number of things with gcd of 1 using inclution exclution
+to do this: sigma_over_possible_j_as_product_of_primes(mu[j] * f(j))
+where f(j) returns number of things where their gcd is divisible by every elemnt of j's factorization, ie. 'j'.

mathematics/number-theory.md

@@ -60,8 +60,12 @@ sieve(10000000); // can go up to 10^7 (need few seconds)
 * O(log10 n): `n = max(a, b)`
 
 ```cpp
-int gcd(int a, int b) {
- return b == 0 ? a : gcd(b, a % b);
+long long gcd(long long a, long long b) {
+ while (a > 0 && b > 0) {
+ if (a > b) a %= b;
+ else b %= a;
+ }
+ return a + b;
 }
 int lcm(int a, int b) {
  return a * (b / gcd(a, b));