onriv

    /**
     * TODO refactor
     */
    @Override
    public XResPostableStatus canXResPostable(XResPostableOrder order) {

        boolean valid = true;
        String msg = null;
        String orderStatus = order.getOrderStatus();
        OrderCategory orderCategory = order.getOrderCategory();

onriv

\documentclass[8pt]{memoir} 
\usepackage[utf8]{inputenc}
\usepackage[paperwidth=9cm, paperheight=11.5cm, top=0.5cm, left=0.5cm, right=0.5cm, bottom=0.5cm]{geometry}
\usepackage{graphicx}
\usepackage{tikz, blindtext}

\makechapterstyle{box}{
  \renewcommand*{\printchaptername}{}
  \renewcommand*{\chapnumfont}{\normalfont\sffamily\huge\bfseries}
  \renewcommand*{\printchapternum}{

onriv

## Ignore Python compiled files
*.pyc

## Ignore vim temporary files
*.swp
*~

## Ignore JetBrains temporary files
.idea/

onriv

#!/usr/bin/env/ python
#coding=utf-8
import mechanize
import cookielib
import itertools
import string
import threading
import time
# Cookie Jar
cj = cookielib.LWPCookieJar()

onriv

    import ConfigParser, sys
    _data = ConfigParser.ConfigParser()
    _data.read(sys.argv[1])

onriv

# -*- coding: utf-8 -*- 

'''
输入一个数组（元素假设为整数），对其元素进行赋值（赋予权重，为整数），使得
 - 每个元素的权重至少为1
 - 如果某一元素比其相邻的元素大（左或右），那么该元素的权重也比对应元素的权重大

输出：权重最小的赋值算法

解决思路：

onriv

#!/usr/bin/env python
# -*- coding: utf-8

"""
Search longest common substrings using generalized suffix trees built with Ukkonen's algorithm

Author: Ilya Stepanov <code at ilyastepanov.com>

(c) 2013
"""

onriv

# -*- coding: utf-8 -*- 

# 依次替换示例
def f(m, count=[0]):
	count[0] += 1
	return str(count[0])

re.sub('a', f, 'ababab')
# 输出为
# '1b2b3b'

onriv

# -*- coding: utf-8 -*- 
import re

def changeText(paragraph, prefix, countMatch = 0):
	pattern = r'[①②③④⑤⑥⑦⑧⑨⑩]'
	text = '<a class="duokan-footnote" href="%s#df-%d"><img src="../Images/note.png"/></a>'
	def changeRule(matchobj):
		nonlocal countMatch
		countMatch += 1
		return text % (prefix, countMatch)

onriv

#!/usr/bin/python -O

# partition A[p:r] in place about x, and return the final position of the pivot
def partition(A, p, r, x):
  # find the index of the pivot
  i = -1
  for j in range(p, r):
    if A[j] == x:
      i = j
      break