clementi

(defn factorial [n]
  (if (= n 0)
      1
      (* n (factorial (- n 1)))))

clementi

var prefixes = [
    {
        prefix: "",
        name: ""
    },
    {
        prefix: "Ki",
        name: "kibi"
    },
    {

clementi

var prefixes = [
    {
        prefix: "",
        name: ""
    },
    {
        prefix: "Ki",
        name: "kibi"
    },
    {

clementi

namespace Ncoa.Matching.UI
{
	using System.Collections.Generic;
	using System.Linq;
	using StorageAccess;

	public abstract class FakeQueryStorage<T> : IQueryStorage
	{
		protected IEnumerable<T> items;

clementi

def logistic_map(x0, r, n):
    if n == 0:
        return x0

    x_prev = logistic_map(x0, r, n - 1)
    return r * x_prev * (1 - x_prev)

clementi

var prefixes = [ "", "Ki", "Mi", "Gi", "Ti", "Pi", "Ei" ];

const GROUP = 1024;

function binaryUnitFormat(x, unit) {
    var index = 0;
    var xnew = x;
    
    while (xnew >= GROUP && index < prefixes.length - 1) {
        xnew /= GROUP;

clementi

var Iterable = {
    func: {
        predicate: {
            TRUE: function (x) { return true; }
        }
    }
};

Array.prototype.forEach = function (fn) {
    for (var i = 0; i < this.length; i++)

clementi

var Iterable = {
    func: {
        predicate: {
            TRUE: function (x) { return true; }
        }
    }
};

Array.prototype.forEach = function (fn) {
    for (var i = 0; i < this.length; i++)

clementi

head [ (a, b, c) | c <- [1..], b <- [1..c], a <- [1..b], a^2 + b^2 == c^2, a + b + c == 1000 ]

clementi

from itertools import permutations

MILLIONTH = 10 ** 6 - 1

print reduce(lambda x, y: str(x) + str(y), list(permutations(range(10), 10))[MILLIONTH])