Java-2018-1-9 first,
Today is the first day of the final exam, and the first day of getting started with Java. If you don't want to review it, learn something new !!
Three major Java platforms
1. Java SE
There are four parts: JVM (Virtual Machine), JRE (runtime environment), JDK (Development Kit), Java
2. Java EE
3. Java ME --- (used less)
Java Development Environment Configuration
Notepad/ediplus/ultreadit
Eclipse/MyEclipse
JCreator
Betbea
PyCharm latest 2018 activation code, latest method, pycharm2018 activation code
Content: Modify the hosts file so that pycharm cannot online verify the activation code.
1. Modify the hosts file
File Location: C: \ Windows \ System32 \ drivers \ etc
Add 0.0.0.0 account.jetbrains.com at the end of the file.
2. Open PyCharm and select Activate code (activated with an activation code)
EB101IWSWD-eyJsaWNlbnNlSWQiOiJFQjEwMUlXU1dEIiwibGljZW5zZWVOYW1lIjoibG
of Windows is a terrible thing, and once an attacker has administrator privileges, it can do anything. Windows may modify the registry, steal secret files and so on, while attacking can also hide themselves, modify the directory files to erase their own traces of intrusion. Therefore, in order to avoid the right to be raised, regular patching, upgrading the system, to avoid being the object of attack.Resources
Metasploit under Windows Multiple right-of-way
Msf_bypassuac the right t
edge.Then the test instructions can be converted to a minimum spanning tree of 22 shortest-circuiting as a weighted value for a given $m$ point.Set $p_x$ and $d_x$ respectively to indicate the closest to each point of the given string and corresponding distance, can be found in BFS, then for a side $ (u,v) $, in the spanning tree to add a connection $p_u$ and $p_v$, the weight is $d_u+d_v+1$ edge.Time Complexity $o (N2^n\alpha (m)) $.#include H. MasterpieceIf each point is not considered to h
Title: Give a positive integer n, ask if there is x, satisfy the condition 2^x mod n=1, if present, find the minimum value of x.Analysis: 1, if the given n is 1, then there is certainly no such x;2, if the given is even, 2 of the power to take more than an even number is definitely to the even, so also can not find;3, if the given is an odd number, its digit number is 3, 5, 7, 9, and 2 of the power of the units of 2, 4, 6, 8, respectively-1 is 1, 3, 5, 7, that is, odd multiples can find relative
good after all, so I began to play violence for the third time after brainstorming about an hour later.Unfortunately, I accidentally dropped a 5-minute part of the back.I thought to myself, "is the time after all rubbish time?" I'd like to get a little more points.The first question looks quite good to do, try it.Finally, 30 minutes before the end of the exam, I get to the T1.I excited to open the T2,T3, with the fastest speed of the file read and write, and then decisively threw away the two q
$\color{red}{\mathsf{update}}$: See Li Aling in the column all about Texnique released installation tutorialAbout TeX live:http://tug.org/texliveUninstall old version of TeX LiveSee Https://tex.stackexchange.com/questions/185520/how-uninstall-texlive-2013-under-windows-8Installation method
Installing TeX Live over the Internet
Downloading one huge ISO file
Online installations often fail. Please consider using an ISO installation.How to install TeX Live
Describe:There is a cow, which has a small cow at the beginning of each year. Each heifer starts its fourth year with a heifer. Please program implementation how many cows are there in the first n years? Input: The input data consists of multiple test instances, one for each test instance, including an integer n (0N=0 indicates the end of the input data and does not handle it. Output: For each test instance, output the number of cows in the nth year.Each output occupies one row. input:2450output
(DIS[U][TEMPK]+W//Shortest path on the same floor{DIS[V][TEMPK]=dis[u][tempk]+W; Que.push (Qnode (V,TEMPK,DIS[V][TEMPK)); } if(tempkk) {if(dis[u][tempk]1])//If this edge value is changed to 0, it will enter the TEMPK+1 layer.{DIS[V][TEMPK+1]=DIS[U][TEMPK]; Que.push (Qnode (V,TEMPK+1, dis[v][tempk+1])); } } } }}intMain () {intT; scanf ("%d",T); while(t--) {scanf ("%d%d%d",n,m,k); Init (); for(intI=0; i) { intu,v,w; scanf ("%d%d
enable the compiled executable document to be debugged with GDB
New exploit.c, code below, \x?? \x?? \x?? \x?? Need to add shellcode to the address stored in memory because the location can overwrite the return address just after an overflow occurs.
We want to get shellcode in-memory address, enter commands gdb stack anddisass main
According to strcpy(buffer + 100,shellcode) the statement, we calculate shellcode the address as0xffffd350(十六进制) + 0x64(100的十六进制) = 0xffffd3b4(十六进制)
Mo
1. Machine-Level Code(1) Two kinds of abstract
Defines the format and behavior of machine-level programs by ISA
The memory address used by the machine-level program is the virtual address
2. Data format3. Operand designator4. Press in and eject stack data
Follow the principle of first in and out
Push Press in, pop delete
Pushq press four words into the stack popq four words pop-up stack
5. Arithmetic and logical operations
LEAQ Load Valid address
INC plus a
D
2018-2019-1 20165334 "Fundamentals of Information Security system Design" Third week study summary and Buffer Overflow Vulnerability experiment One, instruction learning gcc -Og -o xxx.c learns to -Og tell the compiler to use an optimization level that generates machine code that conforms to the overall structure of the original C language code. gcc -Og -S xxx.cLearning ( -S option to view compiled code generated by the C language compiler) gcc -Og -c
T1 (binary search)Test instructionsGiven a sequence, the number of non-empty sets, which can be divided into equal two copies, is obtained.namely [Usaco2012 open]balanced Cow subsets;Data range NTransform the model to know that there are only 3 cases for each number,1, not be selected;2, is selected into the left set;3, is selected to the right set.If the violence enumerates all the possible 3^n will obviously time out,So we can use meet in the middle idea, namely binary search.Enumerate to the
TopicA. Elevator or stairs?DescriptionMasha to go from the X-storey to the Y-floor to find Egor, you can choose to climb stairs or take a helicopter elevator. It is known that climbing stairs each layer needs time T1; The helicopter elevator each floor needs time T2, the helicopter elevator opens or closes once needs the time T3, the current helicopter elevator in the z floor, the helicopter elevator door is in the closed state. If the total time to climb stairs is strictly less than the helicop
non-gate or non-gate
HCL integer Expression
Case Expression Format:
[ select 1: expr 1 select 2: expr 2 . select k: expr k ]
Set Relationship:iexp in{ iexp1,iexp2,...iexpk }
Arithmetic/logic unit (ALU)
Sequential implementation of Y86-64
Organize the processing into stages
Value fetch--> decoding decode--> performing execute--> memory--> writeback write back write back--> update PC update
SEQ
week's exam error summary 1.The following jump commands are related to ZF ()A. jmpB. JeC. jsD. JaE. JBF. JbeAnalytical:2.Assuming that the function of the C-expression T=a+b is completed with the add instruction, the correct statement about the condition Code Register is ()A. If t==0, then zf=1B. If tC. If tD. if (aE. if (aF. LEAQ directive does not affect the condition code registerG. CMP directives do not affect the condition code registerAnalysis: Textbook p135ZF: 0 logo. The result of the r
I. Learning Objectives
Understanding the role of ISA abstraction
Master Isa, and be able to learn other architecture extrapolate
Understanding the pipeline and how it is implemented
Second, the Learning content y86-64 directive
MOVQ directive IRMOVQ rrmovq mrmovq RMMOVQ
Four integer manipulation instructions Addq,subq,andq,xorq only the Register data
7 Jump Instructions Cmovle cmovl cmove cmovne cmovge CMOVG
The call command returns the address to the stack, and then j
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