jeremiahbiard

return newReleases.filter(function(movie) {
    return (movie.rating === 5.0);
  }).map(function(movie) {
    return movie.id;
  });

jeremiahbiard

function complement(A, B) {
    return A.filter(function(elem) { return B.indexOf(elem) == -1; });
}

function unique(arr) {
  return arr.filter(function(elem, pos) {
    return arr.indexOf(elem) == pos;
  });
}

jeremiahbiard

var f = [];
function factorial (n) {
  if (n == 0 || n == 1)
    return 1;
  if (f[n] > 0)
    return f[n];
  return f[n] = factorial(n-1) * n;
}

jeremiahbiard

Twitch API Zipline

A Pen by Jeremiah Biard on CodePen.

License.

jeremiahbiard

int fast_pow(long long base, long long n,long long M)
{
    if(n==0)
       return 1;
    if(n==1)
    return base;
    long long halfn=fast_pow(base,n/2,M);
    if(n%2==0)
        return ( halfn * halfn ) % M;
    else

jeremiahbiard

Beginner: Linked List, Stack, Queue, Binary Search Tree.
 
Intermediate: Heap, Priority Queue, Huffman Tree, Union Find, Tries, Hash Table, Tree Map.

Proficient: Segment Tree, Binary Indexed Tree, Suffix Array, Sparse Table, Lowest Common Ancestor, Range Tree.

Expert: Suffix Automaton, Suffix Tree, Heavy-Light Decomposition, Treap, Aho-Corasick, K-d tree, Link-Cut Tree, Splay Tree, Palindromic Tree, Rope,  Dancing Links, Radix Tree, Dynamic Suffix Array.

jeremiahbiard

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <map>
using namespace std;

long long degree(long long a, long long k, long long p) {

jeremiahbiard

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <map>
using namespace std;

long long degree(long long a, long long k, long long p) {

jeremiahbiard

`The required answer is : 1 / (n + k) * n-choose-r(n - 3,k) * n-choose-r(n + k, n - 1).

While I am not aware of a simple way to prove this which I can describe here, here's some practical advice on how such problems can be approached. It helps to write down a simpler solution and searching Google or OEIS (Online Encyclopedia of Integer Sequences) for small inputs like k = 3 and N = 1,2,...10. This usually gives you a formula describing the sequence, and possibly generalizations for the same.

Computing the answer using the above formula requires calculating the power of MOD (1e6 + 3) in the numerator and the demonimator. If the power of MOD is bigger in the numerator, the answer is 0. Otherwise, we can calculate all the factorials with the powers of MOD cast out. For example, if MOD = 3 and N = 11, we would calculate the product 1 * 2 * 3 / 3 * 4 * 5 * (6 / 3) * 7 * 8 * (9 / 9) * 10 * 11. This can be done easily in O(MOD). Once we have that, we can multiply the values in the numerator and mltiply with the mo

jeremiahbiard

start =
	expression
    
validchar
    = [0-9a-zA-Z_?!+\-=@#$%^&*/.]

_ =
	" "*

list