neesenk

#include <stdio.h>
#include <stdlib.h>
#include <sys/mman.h>

struct env {
  int x;
};

struct __attribute__((packed)) thunk {
  unsigned char push;

neesenk

#include <assert.h>
#define CMP(a, b) (a > b ? 1 : (a == b ? 0 : -1))
int bsearch(const int arr[], int n, int k)
{
	int l = 0, m, cmp;
	assert(n > 0);
	while (l < n) {
		m = (l + n) / 2;
		cmp = CMP(k, arr[m]);
		if (cmp == 0)

neesenk

#include <stdint.h>
#include <stddef.h>

// 0-9999对应的字符串
static const char perfmtstr[] = /* 40000 字节 */
"0000000100020003000400050006000700080009001000110012001300140015001600170018001900200021002200230024"
"0025002600270028002900300031003200330034003500360037003800390040004100420043004400450046004700480049"
"0050005100520053005400550056005700580059006000610062006300640065006600670068006900700071007200730074"
"0075007600770078007900800081008200830084008500860087008800890090009100920093009400950096009700980099"
"0100010101020103010401050106010701080109011001110112011301140115011601170118011901200121012201230124"

neesenk

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>

/**
 * Jaro–Winkler distance
 * http://en.wikipedia.org/wiki/Jaro%E2%80%93Winkler_distance
 *
 * 比较两个字符串的相似程度

neesenk

/*
 * An algorithm for solving the following (classic) hard interview problem:
 *
 * "You are given an array of integers of length n, where each element ranges
 *  from 0 to n - 2, inclusive.  Prove that at least one  duplicate element must
 *  exist, and give an O(n)-time, O(1)-space algorithm for finding some
 *  duplicated element.  You must not modify the array elements during this
 *  process."
 *
 * This problem (reportedly) took CS legend Don Knuth twenty-four hours to solve

neesenk

/**
 * 异步加载js, 支持依赖关系
 * Author: zhiyongLiu <zhiyongliu@tencent.com>
 *
 * 使用方法: LazyLoader().load('a.js').load('b.js').wait().load('c.js')
 *                       .next()
 *                       .load('d.js').load('e.js').wait().load('f.js');
 * c.js 依赖与a.js,b.js, f.js 依赖于d.js,e.js; 这两组js之间没有依赖关系;
 * c.js将在a.js和b.js之后加载, f.js将在d.js,e.js之后加载; 两组js加载同时进行;
 */

neesenk

/*! Copyright (c) 2011 Brandon Aaron (http://brandonaaron.net)
* Licensed under the MIT License (LICENSE.txt).
*
* Thanks to: http://adomas.org/javascript-mouse-wheel/ for some pointers.
* Thanks to: Mathias Bank(http://www.mathias-bank.de) for a scope bug fix.
* Thanks to: Seamus Leahy for adding deltaX and deltaY
*
* Version: 3.0.6
*
* Requires: 1.2.2+

neesenk

/**
 * zhiyongliu <zhiyongliu@tencent.com>
 *
 * 查找一个字符串的所有的重复字串
 * 算法的核心思想是两个相同字符串的前缀一定相同, 在查找过程中先查找短的重复字串,
 * 然后利用短的字串构建次短的重复字串, 直到字符串本身的长度或前一个结果为空集
 * 算法的复杂度为O(n) + O(m), n为字符串长度, m为重复字串的数目
 */
function findRepeatedSubstring(str)
{

neesenk

/**
 * zhiyongliu <zhiyongliu@tencent.com>
 *
 * conf.getData(str, callback) 用来获取数据, 获取的数据通过回调传入, callback传入的参数要是一个数组
 * conf.createChild(obj, inputvalue) 用来生成补全的节点, obj为回调callback的数组的某一项,
 *      在数组为空时obj可能为null,生成出错信息, inputval时当前的input的输入值;
 *      返回一个dom对象，将会作为一个子节点插入到rootid节点中;
 *      函数根据需要可能要设置mouseover/mouserout/click的事件处理函数
 * conf.toSting(obj) 返回obj对应的字符串, obj为回调callback的数组的某一项
 * conf.enterAction(inputvaule) input的回车时的动作可选

neesenk

/**
 * js前端的模版
 * zhiyongliu <zhiyongliu@tencent.com>
 *
 * 支持两种指令: <!-- -->, <!--{}-->
 * 1. <!-- --> js的控制结构的代码；不包含输出
 * 2. <!--{}--> 输出的内容, 结果为string或可转换为string的表达式
 */
var MyTemplate = MyTemplate || function(id) {
        var analyze = function (str) {