Lecture 5



Lecture 5的更多相关文章
- [C2P3] Andrew Ng - Machine Learning
##Advice for Applying Machine Learning Applying machine learning in practice is not always straightf ...
- note of introduction of Algorithms(Lecture 3 - Part1)
Lecture 3(part 1) Divide and conquer 1. the general paradim of algrithm as bellow: 1. divide the pro ...
- codeforces 499B.Lecture 解题报告
题目链接:http://codeforces.com/problemset/problem/499/B 题目意思:给出两种语言下 m 个单词表(word1, word2)的一一对应,以及 profes ...
- Nobel Lecture, December 12, 1929 Thermionic phenomena and the laws which govern them
http://www.nobelprize.org/nobel_prizes/physics/laureates/1928/richardson-lecture.pdf OWEN W. RICHARD ...
- Jordan Lecture Note-1: Introduction
Jordan Lecture Note-1: Introduction 第一部分要整理的是Jordan的讲义,这份讲义是我刚进实验室时我们老师给我的第一个任务,要求我把讲义上的知识扩充出去,然后每周都 ...
- Jordan Lecture Note-3: 梯度投影法
Jordan Lecture Note-3:梯度投影法 在这一节,我们介绍如何用梯度投影法来解如下的优化问题: \begin{align} \mathop{\min}&\quad f(x)\n ...
- Jordan Lecture Note-2: Maximal Margin Classifier
Maximal Margin Classifier Logistic Regression 与 SVM 思路的不同点:logistic regression强调所有点尽可能远离中间的那条分割线,而SV ...
- [CF Round #294 div2] E. A and B and Lecture Rooms 【树上倍增】
题目链接:E. A and B and Lecture Rooms 题目大意 给定一颗节点数10^5的树,有10^5个询问,每次询问树上到xi, yi这两个点距离相等的点有多少个. 题目分析 若 x= ...
- Codeforces Round #287 D.The Maths Lecture
The Maths Lecture 题意:求存在后缀Si mod k =0,的n位数的数目.(n <=1000,k<=100); 用f[i][j]代表 长为i位,模k等于j的数的个数. 可 ...
- Lecture Halls
Lecture Halls (会议安排) 时间限制(普通/Java):1000MS/10000MS 运行内存限制:65536KByte 总提交: 38 测试通过: 2 ...
随机推荐
- NumPy 从已有的数组创建数组
NumPy 从已有的数组创建数组 本章节我们将学习如何从已有的数组创建数组. numpy.asarray numpy.asarray 类似 numpy.array,但 numpy.asarray 只有 ...
- Maximum Gap (ARRAY - SORT)
QUESTION Given an unsorted array, find the maximum difference between the successive elements in its ...
- Codeforces Beta Round #55 (Div. 2)
Codeforces Beta Round #55 (Div. 2) http://codeforces.com/contest/59 A #include<bits/stdc++.h> ...
- TDD - 登录成功和失败
/** * Created by Administrator on 2017-04-06. */ @RunWith(SpringJUnit4ClassRunner.class)@SpringBootT ...
- c#发送短信
短息计费平台:http://sms.webchinese.cn/User/?action=key 代码: using System;using System.Collections.Generic;u ...
- cf相关命令
进行登录的命令: cf login -a https://api.bupaas.citicsinfo.com --skip-ssl-validation 进行发布的命令: cf push gwdemo ...
- Non-negative Integers without Consecutive Ones
n位二进制,求不包含连续1的二进制(n位)数字个数. http://www.geeksforgeeks.org/count-number-binary-strings-without-consecut ...
- [z] Git pull 强制覆盖本地文件
git fetch --all git reset --hard origin/master git pull git fetch 只是下载远程的库的内容,不做任何的合并 git reset 把HEA ...
- python string tuple list dict 相互转换的方法
dict = {'name': 'Zara', 'age': 7, 'class': 'First'}# 字典转为字符串,返回:<type 'str'> {'age': 7, 'name' ...
- (转)ScriptManager.RegisterStartupScript方法和Page.ClientScript.RegisterStartupScript() 方法
ScriptManager.RegisterStartupScript方法 如果页面中不用Ajax,cs中运行某段js代码方式可以是: Page.ClientScript.RegisterStartu ...