Have you ever seen players eyes light up as they explore the worlds that you ve created in your games? If you have, then game development probably has you hooked firmly in its grasp! If you ve never taken your games beyond the PC, now s the time! "J2ME Game Programming" is a hands-on guide that teaches you how to create games for micro-devices. You ll be amazed at just how cool the games you create can look and play. Focusing primarily on mobile phone game creation, you ll jump right in and create your own games as you work your way through the book. The thought has surely crossed your mind that it would be nice to make some money off of this cool hobby of yours. J2ME offers real opportunity to profit from your games. Learn how you can earn revenue from your games by taking them to market. If you have a basic understanding of Java, then you re ready to explore all that "J2ME Game Programming" has to offer!
标签： you created players explore
上传时间： 2017-02-25
上传用户：lanhuaying
在网络安全中经常会遇到rootkit，NSA安全和入侵检测术语字典( NSA Glossary of Terms Used in Security and Intrusion Detection)对rootkit的定义如下：A hacker security tool that captures passwords and message traffic to and from a computer. A collection of tools that allows a hacker to provide a backdoor into a system, collect information on other systems on the network,mask the fact that the system is compromised, and much more. Rootkit is a classic example of Trojan Horse software. Rootkit is available for a wide range of operating systems.
上传时间： 2017-04-04
上传用户：xaijhqx
Playfair Cipher 1.not even the large number of keys in a monoalphabetic cipher provides security 2.one approach to improving security was to encrypt multiple letters 3.the Playfair Cipher is an example 4.invented by Charles Wheatstone in 1854,but named after his friend Baron Playfair Playfair Key Matrix 1.a 5X5 matrix of letters based on a keyword 2.fill in letters of keyword (sans duplicates) 3.fill rest of matrix with other letters Encrypting and Decrypting -plaintext is encrypted two letters at a time 1. if a pair is a repeated letter, insert filler like X’ 2. if both letters fall in the same row, replace each with letter to right (wrapping back to start from end) 3. if both letters fall in the same column, replace each with the letter below it (again wrapping to top from bottom) 4. otherwise each letter is replaced by the letter in the same row and in the column of the other letter of the pair
标签： monoalphabetic Playfair provides security
上传时间： 2017-05-25
上传用户：变形金刚
将魔王的语言抽象为人类的语言：魔王语言由以下两种规则由人的语言逐步抽象上去的：α-〉β1β2β3…βm ；θδ1δ2…-〉θδnθδn-1…θδ1 设大写字母表示魔王的语言，小写字母表示人的语言B-〉tAdA，A-〉sae，eg：B(ehnxgz)B解释为tsaedsaeezegexenehetsaedsae对应的话是：“天上一只鹅地上一只鹅鹅追鹅赶鹅下鹅蛋鹅恨鹅天上一只鹅地上一只鹅”。(t-天d-地s-上a-一只e-鹅z-追g-赶x-下n-蛋h-恨)
上传时间： 2013-12-19
上传用户：aix008
A user-space device driver can do many of the things that kernel drivers can t, such as perform a long-running computation, block while waiting for an event, or read files from the file system. Unlike kernel drivers, a user-space device driver can use other device drivers--that is, access the network, talk to a serial port, get interactive input from the user, pop up GUI windows, or read from disks. User-space drivers implemented using FUSD can be much easier to debug it is impossible for them to crash the machine, are easily traceable using tools such as gdb, and can be killed and restarted without rebooting even if they become corrupted. FUSD drivers don t have to be in C--Perl, Python, or any other language that knows how to read from and write to a file descriptor can work with FUSD. User-space drivers can be swapped out, whereas kernel drivers lock physical memory.
标签： user-space can drivers perform
上传时间： 2014-01-01
上传用户：saharawalker
本代码为编码开关代码，编码开关也就是数字音响中的 360度旋转的数字音量以及显示器上用的（单键飞梭开 关）等类似鼠标滚轮的手动计数输入设备。 我使用的编码开关为5个引脚的，其中2个引脚为按下 转轮开关（也就相当于鼠标中键）。另外3个引脚用来 检测旋转方向以及旋转步数的检测端。引脚分别为a,b,c b接地a，c分别接到P2.0和P2.1口并分别接两个10K上拉 电阻，并且a，c需要分别对地接一个104的电容，否则 因为编码开关的触点抖动会引起轻微误动作。本程序不 使用定时器，不占用中断，不使用延时代码，并对每个 细分步数进行判断，避免一切误动作，性能超级稳定。 我使用的编码器是APLS的EC11B可以参照附件的时序图 编码器控制流水灯最能说明问题，下面是以一段流水 灯来演示。
上传时间： 2017-07-03
上传用户：gaojiao1999
【问题描述】 在一个N*N的点阵中，如N=4,你现在站在（1，1），出口在（4，4）。你可以通过上、下、左、右四种移动方法，在迷宫内行走，但是同一个位置不可以访问两次，亦不可以越界。表格最上面的一行加黑数字A[1..4]分别表示迷宫第I列中需要访问并仅可以访问的格子数。右边一行加下划线数字B[1..4]则表示迷宫第I行需要访问并仅可以访问的格子数。如图中带括号红色数字就是一条符合条件的路线。 给定N，A[1..N] B[1..N]。输出一条符合条件的路线，若无解，输出NO ANSWER。（使用U，D，L，R分别表示上、下、左、右。） 2 2 1 2 （4，4） 1 （2，3） （3，3） （4，3） 3 （1，2） （2，2） 2 （1，1） 1 【输入格式】 第一行是数m (n < 6 )。第二行有n个数，表示a[1]..a[n]。第三行有n个数，表示b[1]..b[n]。 【输出格式】 仅有一行。若有解则输出一条可行路线，否则输出“NO ANSWER”。
标签： 点阵
上传时间： 2014-06-21
上传用户：llandlu
实验源代码 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("请输入矩阵第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可传递闭包关系矩阵是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元关系的可传递闭包\n"); void warshall(int,int); int k , n; printf("请输入矩阵的行数 i: "); scanf("%d",&k); 四川大学实验报告 printf("请输入矩阵的列数 j: "); scanf("%d",&n); warshall(k,n); }
上传时间： 2016-06-27
上传用户：梁雪文以
#include "iostream" using namespace std; class Matrix { private: double** A; //矩阵A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //为向量b分配空间并初始化为0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //为向量A分配空间并初始化为0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析构中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"请输入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"请输入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"个:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分别求得U，L的第一行与第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分别求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"计算U得:"<<endl; U.Disp(); cout<<"计算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
标签： 道理特分解法
上传时间： 2018-05-20
上传用户：Aa123456789
The telecommunications industry is undoubtedly in a period of radical change with the advent of mobile broadband radio access and the rapid convergence of Internet and mobile services. Some of these changes have been enabled by a fundamental shift in the underlying technologies; mobile networks are now increasingly based on a pure Internet Protocol (IP) network architecture. Since the first edition of this book was published in 2009, a multitude of connected devices from eBook readers to smartphones and even Machine-to-Machine (M2M) technologies have all started to benefit from mobile broadband. The sea change over the last few years is only the beginning of a wave of new services that will fundamentally change our economy, our society, and even our environment. The evolution towards mobile broadband is one of the core underlying parts of this revolution and is the focus of this book.
标签： Networks Packet EPC and 4G
上传时间： 2020-05-27
上传用户：shancjb
