mebusw

# !encoding=utf8
# primes = filterPrime [2..] 
#   where filterPrime (p:xs) = 
#           p : filterPrime [x | x <- xs, x `mod` p /= 0]

# Infinite List
from pipe import *
from pipe import itertools

def primes(xs):

mebusw

# https://codingstyle.cn/topics/202

def f(n, p=2):
    if n < 2: return []
    if n % p == 0: return [p] + f(n/p, p)
    return f(n, p+1)
    
assert [2,2,3,3,3,5,5,19] == f(2*2*3*3*3*5*5*19)

mebusw

#! /usr/local/bin/python
# -*- coding: utf-8 -*-
import subprocess, time

# 程序执行终止时间为当前时刻延迟N秒
stoptime = time.strftime('%Y-%m-%d %H:%M:%S', time.localtime(time.time()+60))


def main(command):
    popen = subprocess.Popen(

mebusw

import random

TIMES = 10000

def once(acc=0, days=0):
	return days if acc >= 20 else once(acc + random.choice(xrange(1,7)), days + 1)
	

s1 = [once() for i in xrange(TIMES)]
s1.sort()

mebusw

def balance_recurse(chars, cnt=0):
	if len(chars) == 0: return cnt == 0
	if chars[0] == '(': cnt += 1
	if chars[0] == ')': cnt -= 1
	return cnt >= 0 and balance(chars[1:], cnt)

PARENS = (('(', ')'), ('[', ']'),)
def balance_recurse_mul(chars, cnt=[0]*len(PARENS)):
	if len(chars) == 0: return all(map(lambda b: b==0, cnt))
	for i, paren in enumerate(PARENS):

mebusw

#!/usr/local/bin/python
# -*- coding: UTF-8 -*-
import numpy as np
import matplotlib.pyplot as plt


def predict(input_vec, f):
    # print (input_vec, W)
    return f(np.dot(input_vec, W))

mebusw

import scrapy

class BookSpider(scrapy.Spider):
    name = "books"
    start_urls = ['http://books.toscrape.com/']

    def parse(self, response):
        for book in response.css('article.product_pod'):
            name = book.xpath('./h3/a/@title').extract_first()
            price = book.css('p.price_color::text').extract_first()

mebusw

import scrapy

class CsxCourseSpider(scrapy.Spider):
    name = "csx"
    start_urls = ['https://www.scrumalliance.org/courses-events/course.aspx?pageCount=50&country=&state=&city=&zip=&type=Csd;&trainer=&language=&startdate=5/15/2017%2012:00:00%20AM&enddate=1/1/1900%2012:00:00%20AM&discount=False&page=1&orderby=StartDate&sortdir=asc&radius=0&view=map']

    # https://www.scrumalliance.org/courses-events/courses/csd/us/ohio/columbus/2017/may/201702145-csd

    def parse(self, response):
        # print '}}}}}}}}', response.url, response.xpath('*//tr')

mebusw

#include "include.hpp"
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
using std::cout;
using std::endl;
using std::string;
using std::ifstream;
using std::ofstream;

mebusw

#!encoding=utf8
"""
建立在山峰和峡谷的城堡酒店，看具体描述：https://coding.net/u/wanghao/p/The-Castle-Company/git/blob/master/README.md

建立在山峰和峡谷的城堡酒店

给定一组整数数组，可以把它们想象成高度，如［3，2，5，6，7，5，3］从而形成一条地形数据。

要求：找到所有的峰值和谷值的总数，所谓峰值就是他之前的数字和之后的数字都比他小，谷值就是之前和之后的数字都比他大。
如 ［2，4，3］中 4 可以记为一个峰值， ［5，3，6］中3可以记为一个谷值，这个峰／谷值可以是一个数或者一系列相同的数，如 ［3，5，5，5，4］ 三个 5 可以看作是一个峰值。