Question There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more…

LeetCode算法题目解答汇总 本文转自<四火的唠叨> 只要不是特别忙或者特别不方便,最近一直保持着每天做几道算法题的规律,到后来随着难度的增加,每天做的题目越来越少.我的初衷就是练习,因为一方面我本身算法基础并不好,再一方面是因为工作以后传统意义上所谓算法的东西接触还是太少.为了题目查找方便起见,我把之前几篇陆陆续续贴出来的我对LeetCode上面算法题的解答汇总在下面,CTRL+F就可以比较方便地找到.由于LeetCode上的题在不断更新,因此我也会不定期地更新.下面表格里面的Accep…

There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more candies…

转自:http://www.cnblogs.com/lanxuezaipiao/p/3371224.html 都 是一些非常非常基础的题,是我最近参加各大IT公司笔试后靠记忆记下来的,经过整理献给与我一样参加各大IT校园招聘的同学们,纯考Java基础功底,老 手们就不用进来了,免得笑话我们这些未出校门的孩纸们,但是IT公司就喜欢考这些基础的东西,所以为了能进大公司就~~~当复习期末考吧.花了不少时间整理,在整理过程中也学到了很多东西,请大家认真对待每一题~~~   下面都是我自己的答案非官方,仅…

精通Web Analytics 2.0 : 用户中心科学与在线统计艺术 第六章:使用定性数据解答"为什么"的谜团 当我走进一家超市,我不希望员工会认出我或重新为我布置商店. 然而,当我访问一个在线超市,我很郁闷的是在我第三次访问时,他们仍然不知道我住在加利福尼亚州,他们没有给我介绍在我的本地商店有售的商品. 当人们在网上购物时,他们会有不同的一些期待. 因此,您的Web Analytics 2.0策略必须包括至少几个积极地倾听客户的声音的方法. 通过这种方式,您站在了他们的期望顶端,您…

一.前言 继上一篇写完字节编码内容后,现在分析在Java中各字符编码的问题,并且由这个问题,也引出了一个更有意思的问题,笔者也还没有找到这个问题的答案.也希望各位园友指点指点. 二.Java字符编码 直接上代码进行分析似乎更有感觉. public class Test { public static String stringInfo(String str, String code) throws Exception { byte[] bytes = null; if (code.equals(…

这里贴上applicationContext里的代码: <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="h…

首先声明一下,这个解答是从Segmentfault看到的,挺有意思就记录下来.我放到最下面: bind() https://developer.mozilla.org/zh-CN/docs/Web/JavaScript/Reference/Global_Objects/Function/bind Function.prototype.bind 这个方法是 ECMAScript 5 新增加的,在 Firefox 4/Chrome 中都支持,IE8 应该还不支持. 简介: 创建一个 Function…

There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more candies…

135. Candy There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get mo…

Candy There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more ca…

这五个问题相当经典而且比较深,需要做过CMMI4.5级的朋友才能看懂这些问题.这5个问题是一位正在实践CMMI4级的朋友提出来的,而解答则是我的个人见解. 五个疑问是:   A.流程,子流程部分不明白 如果刚开始做4级,需要把指标与流程对应上,我们应该怎么做呢?是凭经验感觉还是有相关的过程步骤?如果我们凭经验按上面的例子把流程,子流程,子子流程和指标对应了起来,然后通过改善相关流程来提高指标但是如果我们的感觉偏差,对应错了流程和指标的关系,岂不是白费力气? B.组织级和项目级的区别在哪里,如果P…

1. 设函数 $f(x) = 2^x(ax^2 + bx + c)$ 满足等式 $f(x+1) - f(x) = 2^x\cdot x^2$, 求 $f(1)$. 解答: 由 $f(x) = 2^x(ax^2 + bx + c)$, 可以得出 $$f(x+1) = 2^{x+1}[a(x+1)^2 + b(x+1) + c]= 2\cdot2^x[(ax^2 + bx + c) + 2ax + a+ b]= 2\cdot f(x) + 2\cdot2^x\cdot(2ax + a + b)$$…

There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more candies…

很多疑惑一扫而空.... http://www.zhihu.com/question/35905242?sort=created JS的单线程,浏览器的多进程,与CPU,OS的对位. 互联网移动的起起落落... 爽!!! 作者:igetit链接:http://www.zhihu.com/question/35905242/answer/65974599来源:知乎著作权归作者所有,转载请联系作者获得授权. ### 第1个问题:为什么浏览器的开发语言是JavaScript? 因为JavaScript…

题目描述: There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more ca…

C. Inna and Candy Boxes   Inna loves sweets very much. She has n closed present boxes lines up in a row in front of her. Each of these boxes contains either a candy (Dima's work) or nothing (Sereja's work). Let's assume that the boxes are numbered fr…

There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more candies…

这道题只要英语单词都认得,阅读没有问题,就做得出来. POJ - 1666 Candy Sharing Game Time Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %I64d & %I64u Description A number of students sit in a circle facing their teacher in the center. Each student initially has an even n…

[问题2014A01] 解答一(第一列拆分法,由张钧瑞同学提供) (1)  当 \(a=0\) 时,这是高代书复习题一第 33 题,可用升阶法和 Vander Monde 行列式来求解,其结果为 \[|A|=\prod_{1\leq i<j\leq n}(x_j-x_i)\Big(\sum_{i=1}^nx_1\cdots\hat{x}_i\cdots x_n\Big),\] 其中 \(\hat{x}_i\) 表示 \(x_i\) 不在其中. (2)  当 \(a\neq 0\) 时,我们有 \…

[问题2014A01] 解答二(后 n-1 列拆分法,由郭昱君同学提供) \[|A|=\begin{vmatrix} 1 & x_1^2-ax_1 & x_1^3-ax_1^2 & \cdots & x_1^n-ax_1^{n-1} \\ 1 & x_2^2-ax_2 & x_2^3-ax_2^2 & \cdots & x_2^n-ax_2^{n-1} \\ \vdots & \vdots & \vdots & \vd…

[问题2014A01] 解答三(升阶法,由董麒麟同学提供) 引入变量 \(y\),将 \(|A|\) 升阶,考虑如下行列式: \[|B|=\begin{vmatrix} 1 & x_1-a & x_1(x_1-a) & x_1^2(x_1-a) & \cdots & x_1^{n-1}(x_1-a) \\ 1 & x_2-a & x_2(x_2-a) & x_2^2(x_2-a) & \cdots & x_2^{n-1}(x_…

[问题2014A02] 解答一(两次升阶法,由张钧瑞同学.董麒麟同学提供) 将原行列式 \(|A|\) 升阶,考虑如下 \(n+1\) 阶行列式: \[|B|=\begin{vmatrix} 1 & -a_1 & -a_2 & \cdots & -a_{n-1} & -a_n \\ 0 & 0 & a_1+a_2 & \cdots & a_1+a_{n-1} & a_1+a_n \\ 0 & a_2+a_1 &…

[问题2014A02] 解答二(求和法+拆分法,由张诚纯同学提供) 将行列式 \(|A|\) 的第二列,\(\cdots\),第 \(n\) 列全部加到第一列,可得 \[ |A|=\begin{vmatrix} \sum_{i=1}^na_i+(n-2)a_1 & a_1+a_2 & \cdots & a_1+a_{n-1} & a_1+a_n \\ \sum_{i=1}^na_i+(n-2)a_2 & 0 & \cdots & a_2+a_{n-1…

[问题2014A02] 解答三(降阶公式法) 将矩阵 \(A\) 写成如下形式: \[A=\begin{pmatrix} -2a_1 & 0 & \cdots & 0 & 0 \\ 0 & -2a_2 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & -2a_{n-1} &…

[问题2014A03]  解答 注意到 \((A^*)^*\) 的第 (1,1) 元素是 \(A^*\) 的第 (1,1) 元素的代数余子式, 即为 \[\begin{vmatrix} A_{22} & A_{32} & \cdots & A_{n2} \\ A_{23} & A_{33} & \cdots & A_{n3} \\ \vdots & \vdots & \vdots & \vdots \\ A_{2n} & A_…

[问题2014A04]  解答 (1) 由条件可得 \(AB+BA=0\), 即 \(AB=-BA\), 因此 \[AB=A^2B=A(AB)=A(-BA)=-(AB)A=-(-BA)A=BA^2=BA,\] 从而 \(AB=BA=0\). (2) 由条件可得 \(0=B(AB)^kA=(BA)^{k+1}\), 因此 \[(I_n-BA)\Big(I_n+BA+\cdots+(BA)^k\Big)=I_n,\] 从而 \(I_n-BA\) 可逆. (3) 我们给出此小题的三种解法. 解法一(凑…

[问题2014A05]  解答 (1) 将矩阵 \(A\) 分解为两个矩阵的乘积: \[A=\begin{bmatrix} 1 & 1 & \cdots & 1 & 1 \\ x_1 & x_2 & \cdots & x_n & x \\ \vdots & \vdots &  & \vdots & \vdots \\  x_1^{n-1} & x_2^{n-1} & \cdots & x…

[问题2014A06]  解答 用反证法, 设存在 \(n\) 阶正交阵 \(A,B\), 使得 \[A^2=cAB+B^2,\,\,c\neq 0.\cdots(1)\] 在 (1) 式两边同时左乘 \(A'\) 且右乘 \(B'\), 注意到\(A,B\) 都是正交阵, 可得 \[AB'=cI_n+A'B,\] 从而 \[cI_n=A'B-AB'.\cdots(2)\] 在 (2) 式两边同时取迹, 可得 \begin{eqnarray*}nc&=&\mathrm{tr}(cI_n)…

[问题2014A07]  解答 我们分三步进行证明. \(1^\circ\) 先证 \(\alpha_1,\alpha_2\) 线性无关. 用反证法, 设 \(\alpha_1,\alpha_2\) 线性相关, 我们来推出矛盾. 因为 \(\alpha_1\neq 0\), 故 \(\alpha_2\) 可表示为 \(\alpha_1\) 的线性组合, 即存在 \(k\in\mathbb{Q}\), 使得 \(\alpha_2=k\alpha_1\). 带入题中条件可得 \[\alpha_3=k…