na2hiro

function pascal(arr){
	console.log(arr);
	if(arr.length==1) return arr;
	var newarr = [];
	for(var i=0; i<arr.length-1; i++){
		newarr.push((Number(arr[i])+Number(arr[i+1]))%10)
	}
	return pascal(newarr);
}

na2hiro

var Iconv = require("iconv").Iconv, conv = new Iconv('shift_jis','UTF-8//TRANSLIT//IGNORE');
var fs = require("fs");
var filename = process.argv[2];
if(!filename) throw "need argument";

fs.readFile(filename, function (err, data) {
	if (err) throw err;
	var output="";
	conv.convert(data).toString().split("¥n").forEach(function(line){
		var elems = line.split('","');

na2hiro

http://twitter.com/ken5a/status/338877331106050048 "N個の異なる要素から構成される集合{0,1,...,N-1}をK個のグループに分割したい。 注:グループ内での順序やグループ同士の順序はない(つまり{{0,1}, {2, 3}} == {{3,2}, {1, 0}})" "N=3なら [((0, 1, 2),), ((0,), (1, 2)), ((0, 2), (1,)), ((0, 1), (2,)), ((0,), (1,), (2,))]"

• [[0,2],[1],[3]]という分割結果に対し，0→[0,2],1→[1],2→[3]という添字を付け，元の数字の順に添字を並べた0102を添字列と呼ぶ．
• 再帰的に添字列を作る．
• 先頭を0とし，振れる数字を0から(今まで振った数+1)までとすれば，結果の重複は除ける．
  • (例えば添字列0201([[0,2],[3],[1]])は生成されない)
• これは，重複した分割結果[[0,2],[1],[3]]や[[3],[1],[0,2]]などの中で，それぞれの最小の数字が昇順になっている結果だけを生成していることになる．

na2hiro

var script = document.createElement('script');
script.src = 'http://code.jquery.com/jquery-latest.js';
document.getElementsByTagName('head')[0].appendChild(script);

var sessid="~";

function del(name){
  $.post("/index.php/%E7%89%B9%E5%88%A5:%E5%88%A9%E7%94%A8%E8%80%85%E3%81%AE%E7%B5%B1%E5%90%88",{
		olduser:name, 
		newuser:"Anonymous",

na2hiro

var rl = require("readline");
var repl = rl.createInterface(process.stdin, process.stdout);
var input="";
repl.on("line", function(text){
	input+=text;
});
repl.on("close", function(){
	var texts = input.split("\n");
	var t = parseInt(texts[0], 10);
//	console.log(t)

na2hiro

import Data.List.Split(splitOn)
import Data.List(sort, tails, maximumBy)

type Date = (Int, Int)
type Ticket = (Date, Date, String)

parse :: String->Ticket
parse s = ((m1,d1), (m2,d2), name)
    where [name, dates] = words s
          [[m1,d1], [m2,d2]] = map(map read.splitOn "/")$ splitOn "-" dates

na2hiro

定義
(f. g) x = f (g x) -- .
(f. g. h) x = f (g (h x)) -- ..
dual f = dual. f. dual -- dual f
dual xs = reverse (map dual xs) -- dual list
foldr _ z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
reverse [] = []
reverse (x:xs) = reverse xs ++ [x]
map _ [] = []

na2hiro

// 20字(<=100字)
function yourCode() {
    var arr = [];
    for (var i = 1; i <= 100; i ++) {
        arr.push([i, "fizz", "buzz", "fizzbuzz"][i%3?i%5?0:2:i%5?1:3]);
    }
    return arr;
}

na2hiro

import Data.List(sort, nub)

makeNumByN :: [[Integer]]
makeNumByN = [0..9] : map f [0..] -- memoize
  where f k = nub$sort$ do
                l<-[0..k]
                a<-makeNumByN!!l
                b<-makeNumByN!!(k-l)
                [a*b, a+b, if a>b then a-b else b-a] -- negative value has no meaning

na2hiro

IFPH 3.1.4 m*nの評価に必要な評価ステップ数はいくつか．

ただし自然数及び加算，乗算は次の通り．
data Nat = Zero | Succ Nat
(+) :: Nat->Nat->Nat
m + Zero = m
m + Succ n = Succ (m+n)
(*) :: Nat->Nat->Nat
m * Zero = Zero
m * Succ n = (m*n)+m