这是一本计算神经科学的优秀著作，全文拷贝这里（图和公式缺），有兴趣的同学可以
阅读
如需要，我可以分享PDF文件（－－仅供个人学习，无商业用途）
From Computer to Brain
William W. Lytton
From Computer to Brain
Foundations of Computational Neuroscience
Springer
William W. Lytton, M.D.
Associate Professor, State University of New York, Downstato, Brooklyn, NY
Visiting Associate Professor, University of Wisconsin, Madison
Visiting Associate Professor, Polytechnic University, Brooklyn, NY
Staff Neurologist., Kings County Hospital, Brooklyn, NY
In From Computer to Brain: ... 阅读全帖
P0100 Mass or Volume Air Flow Circuit Malfunction
P0101 Mass or Volume Air Flow Circuit Range/Performance Problem
P0102 Mass or Volume Air Flow Circuit Low Input
P0103 Mass or Volume Air Flow Circuit High Input
P0104 Mass or Volume Air Flow Circuit Intermittent
P0105 Manifold Absolute Pressure/Barometric Pressure Circuit Malfunction
P0106 Manifold Absolute Pressure/Barometric Pressure Circuit Range/
Performance Problem
P0107 Manifold Absolute Pressure/Barometric Pressure Circuit Low Input
P0108 ... 阅读全帖
☆─────────────────────────────────────☆
Bayesian1 (Jason) 于 (Tue Aug 30 00:32:06 2011, 美东) 提到:
Given an array A of positive integers. Convert it to a sorted array with
minimum cost. The only valid operation are:
1) Decrement with cost = 1
2) Delete an element completely from the array with cost = value of element
☆─────────────────────────────────────☆
chenpp (chenpp) 于 (Tue Aug 30 00:37:57 2011, 美东) 提到:
my 2 cents:
允许额外花费O(n)空间么。。。
允许的话就不停地减数组中所有元素的值，减一次计数器加1，遇到减到0的就删掉，把
当前计数器值放入新开的等大... 阅读全帖
我的code：
先扫一遍找到A的个数和B的个数
然后再扫一遍处理A
再扫一遍处理B
code里处理A和B的算法不太一样，第二种更好一点。另外，有没有包子？
/*input is assumed to end at '\0' and input is assumed to be long enough to
hold the
* transformed string
* */
static char[] deleteADoubleB(final char[] input) {
if (input == null || input.length == 0) {
return input;
}
int aCount = 0, bCount = 0, originalLen = 0;
for (int i = 0; i < input.length; ++i) {
if (input[i] == 'A') {
aCount ... 阅读全帖
Your input can't be opened:
VLC is unable to open the MRL '1600.m3u8'. Check the log for details.
Your input can't be opened:
VLC is unable to open the MRL '2400.m3u8'. Check the log for details.
Your input can't be opened:
VLC is unable to open the MRL '3000.m3u8'. Check the log for details.
Your input can't be opened:
VLC is unable to open the MRL '4500.m3u8'. Check the log for details.
Your input can't be opened:
VLC is unable to open the MRL 'duh.m3u8'. Check the log for details.
Your input ... 阅读全帖
ok, my bad, keep track of range [0, neg) and [numNeg, pos), and do not
rearrange those (since they are already in the correct place)
use [neg, numNeg) and [pos, N) as circular buffer for un-processed elements
#include
#include
#include
using namespace std;
void print(const vector& input) {
for(size_t i = 0; i < input.size(); ++i) {
cout << input[i] << " ";
}
cout << endl;
}
int numNegatives(const vector& input) {
int cnt = 0;
... 阅读全帖
我已经写了一个，请大家看一下还有什么可以改进的地方
package design_a_pizza_maker;
/*
* @ServiceNow onsite
*/
import java.util.Scanner;
import java.util.Timer;
import java.util.TimerTask;
public class PizzaMaker extends Stopwatch{
public static void main(String[] args) {
PizzaMaker pizzamaker = new PizzaMaker();
Scanner sc = new Scanner(System.in);
System.out.println("Input a to add 30 seconds to current baking
schedule");
System.out.println("Input c to cancel current baking schedule");... 阅读全帖
写了最后那个，没测，大牛测测
void helper(vector > &Input, int i, int j, int n, int m)
{
if(i<0||i>n-1||j<0||j>m-1)
return;
Input[i][j] = -1;
if(i>0&&Input[i-1][j]==1)
helper(Input, i-1, j, n, m);
if(i
helper(Input, i+1, j, n, m);
if(j>0&&Input[i][j-1]==1)
helper(Input, i, j-1, n, m);
if(j
helper(Input, i, j+1, n, m);
return;
}
int GetAdjArea(vector > input)
{
i... 阅读全帖
想出两种思路，都是O(N)的time复杂度。请看注释。
public class HelloWorld{
//Idea：put every element into its right position. which means input[i]
is placed at input[i]-1. Then if input[i]==0, then i+1 is one missing
number.
//this approach modifies the original array.
//O(N) time
public static void printMissing(int[] input){
for(int i=0;i
while(i+1
swap(input, i, input[i]-1);
}
... 阅读全帖
贡献刚做的online test，职位是Machine Learning related。
Question 1 / 2 (LaserMaze)
You are standing in a rectangular room and are about to fire a laser toward
the east wall. Inside the room a certain number of prisms have been placed.
They will alter the direction of the laser beam if it hits them. There
are north-facing, east-facing, west-facing, and south-facing prisms. If the
laser beam strikes an east-facing prism, its course will be altered to be
East, regardless of what direction it had been goi... 阅读全帖
Question 1 / 2 (LaserMaze)
You are standing in a rectangular room and are about to fire a laser toward
the east wall. Inside the room a certain number of prisms have been placed.
They will alter the direction of the laser beam if it hits them. There
are north-facing, east-facing, west-facing, and south-facing prisms. If the
laser beam strikes an east-facing prism, its course will be altered to be
East, regardless of what direction it had been going in before. If it hits
a south-facing prism... 阅读全帖
最终版本的compiler测试结果
=================================================
Output of Building User Code
Explode the tar
c-.l
c-.y
scanType.h
makefile
symtab.h
symtab.cpp
emitCode.h
emitCode.cpp
20131214164956-huang-CS445-F13-A5.tar: POSIX tar archive (GNU)
Tests: directory
c-.l: lex description text
c-.y: lex description text
emitCode.cpp: ASCII C++ program text
emitCode.h: ... 阅读全帖
最终版本的compiler测试结果
=================================================
Output of Building User Code
Explode the tar
c-.l
c-.y
scanType.h
makefile
symtab.h
symtab.cpp
emitCode.h
emitCode.cpp
20131214164956-huang-CS445-F13-A5.tar: POSIX tar archive (GNU)
Tests: directory
c-.l: lex description text
c-.y: lex description text
emitCode.cpp: ASCII C++ program text
emitCode.h: ... 阅读全帖
装了：
Set up Fcitx for Chinese and Japanese language input on Ubuntu Trusty 14.04
fcitx.pngAfter recently upgrading to Ubuntu 14.04 Trusty Tahr (LTS), I
decided to give Fcitx, the default input method framework on Ubuntu’s
Chinese sister project Ubuntu Kylin, a try and I was pleasantly surprised.
Not only is Fcitx rock-solid and actively developed, it also offers input
methods for Japanese, Korean, Vietnamese and a bunch of other languages in
addition to the default Chinese input methods. Here is ... 阅读全帖
用& | 和 bit shift的方法有，只用& |的还真不知道
unsigned __int32 reversing_bits(unsigned __int32 input)
{
// complixity O(log [no.of.bits]) = O(1)
// On 32 bit machines it takes 5 steps (logical)
// Step 1
// Mask bit positions 0, 2, 4... shift LEFT this masked number by one
// Mask bit positions 1, 3, 5... shift RIGHT this masked number by one
input = (input & 0x55555555) << 1 | (input & 0xAAAAAAAA) >> 1;
// Step 2
// Mask bit positions 01, 45, 89... shift LEFT this masked number by ... 阅读全帖
using System;
using System.Text;
using System.Collections.Generic;
public class Calculate
{
private Dictionary ht =new Dictionary();
public Calculate()
{
ht.Add('a', 1);
ht.Add('e', 5);
ht.Add('i', 9);
ht.Add('o', 15);
ht.Add('u', 21);
ht.Add('y', 25);
}
public int Calcu(string input,int start)
{
input=input.ToLower();
int i = start;
while (i < input.Length)
if (!ht.... 阅读全帖
My solution:
1. build a graph by storing "possible next digits" for each digit.
2. dfs each digit in the graph.
#include
#include
using std::vector;
using std::string;
struct Node {
int value;
vector next;
Node(int v) {
value = v;
}
};
void build_tree(vector &input) {
for (unsigned i=0;i
for (unsigned j=i+1;j
if (input[i].value < input[j].value) {
input[i].next.p... 阅读全帖
I am compiling a LaTeX file with 10 chapters and a table of contents. I have
added the following in the first page.
But why is the table of contents still missing?
thanks !!!
Any help will be appreciated.
This is the main file.
\pdfbookmark[1]{TABLE OF CONTENTS}{table}
\tableofcontents
\addtocontents{toc}{\def\protect\@chapapp{}} \cleardoublepage \
phantomsection
\addcontentsline{toc}{chapter}{LIST OF TABLES}
\listoftables
\cleardoublepage \phantomsection \addcontentsline{toc}{chapter}{LIST OF
F... 阅读全帖
I am compiling a LaTeX file with 10 chapters and a table of contents. I have
added the following in the first page.
But why is the table of contents still missing?
thanks !!!
Any help will be appreciated.
This is the main file.
\pdfbookmark[1]{TABLE OF CONTENTS}{table}
\tableofcontents
\addtocontents{toc}{\def\protect\@chapapp{}} \cleardoublepage \
phantomsection
\addcontentsline{toc}{chapter}{LIST OF TABLES}
\listoftables
\cleardoublepage \phantomsection \addcontentsline{toc}{chapter}{LIST OF
F... 阅读全帖
JAVA 不太熟，尝试着快速写了下这两道题，请各位大牛帮忙看看啊，有没有什么
问题。当然我这里没有考虑大数问题，这个呆会我再想想，或者谁教我也好。
public class RomansToInt {
public static HashMap map;
public static int RomanToInt(String input) {
if (input == null || input.length() == 0) {
System.out.println("The input is not valid");
return -1;
}
int len = input.length();
if (len == 1) {
return map.get(input.charAt(0));
}
int result = 0;
int i =... 阅读全帖
for those lazy moms.
You can install iopus on your firefox:
run these codes will add points automatically!
TAG POS=1 TYPE=A ATTR=TXT:GiftstoGrow
TAG POS=1 TYPE=INPUT:TEXT FORM=NAME:quickcode ATTR=ID:redemption_code1
CONTENT=2BEGINEARNING50
TAG POS=11 TYPE=IMG ATTR=SRC:http://en.giftstogrow.pampers.com/static/images/spacer.gif
TAG POS=1 TYPE=A ATTR=TXT:GiftstoGrow
TAG POS=1 TYPE=INPUT:TEXT FORM=NAME:quickcode ATTR=ID:redemption_code1
CONTENT=GIFTSTOGROW4MOM
TAG POS=11 TYPE=IMG ATT... 阅读全帖
也找工作了一段时间了，从版上学了很多，上周G家面完了，求个bless。
之前的一些都挂了，还在继续找其他的。等定下来之后一定发面经回报本版。
谢谢大家啦！！
1. http://www.mitbbs.com/article_t/JobHunting/32005597.html
1) Implement a simple calculator (+,-,*,/);
2) Implement "+1" for a large integer;
3) How to match Ads to users;
4) How to extract useful information from a forum webpage (list all
kinds of useful signal you can think of)
5) How to detect the duplicate HTML pages (large scale);
6) Find all the paths between two places on Google map;
7)... 阅读全帖
Space O(1)
Time O(n)
// Given a string, return the longest substring that contains at most // two
characters.
// solution: scan the string and update the first and second letter's last
occurrence
// indices, and update the solution's start index.
int FindLength(const string& input) {
if (input.size() <= 2) {
return input.size();
}
int solu_start = 0;
int first_end, second_end;
char first = input[0];
char second;
int curr = 1;
while (curr < input.size() && in... 阅读全帖
Space O(1)
Time O(n)
// Given a string, return the longest substring that contains at most // two
characters.
// solution: scan the string and update the first and second letter's last
occurrence
// indices, and update the solution's start index.
int FindLength(const string& input) {
if (input.size() <= 2) {
return input.size();
}
int solu_start = 0;
int first_end, second_end;
char first = input[0];
char second;
int curr = 1;
while (curr < input.size() && in... 阅读全帖
How about this one?
class Interval {
public:
Interval(int start_, int end_): start(start_),end(end_){}
int start;
int end;
};
bool comp(Interval &rhs, Interval &lhs)
{
return (rhs.start == lhs.start ? rhs.end < lhs.end : rhs.start < lhs.start);
}
bool check(vector &input)
{
int left = input[0].start;
int right = input[0].end;
for (int i = 1; i < input.size(); i++) {
if (right < input[i].start) {
return false;
} else {
right = max(input[i].end, right);
}
}
return true;
}
int main()
... 阅读全帖
code assessment是我自己面的。
aptitude paper是从老印论坛扒的。
PROBLEM ONE: TRAINS
Problem: The local commuter railroad services a number of towns in Kiwiland
. Because of monetary concerns, all of the tracks are 'one-way.' That is,
a route from Kaitaia to Invercargill does not imply the existence of a route
from Invercargill to Kaitaia. In fact, even if both of these routes do
happen to exist, they are distinct and are not necessarily the same distance!
The purpose of this problem is to help the railroad p... 阅读全帖
