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[Other] 校验码 DL : 0: 校验码算法，可运行，加说明，例如：代码690123456789X1（共13位）校验码的计算步骤举例说明 1、自右向左编号位置序号 13 12 11 10 9 8 7 6 5 4 3 2 1 代码 6 9 0 1 2 3 4 5 6 7 8 9 X1 2、从序号2开始求出偶数 9+7+5+3+1+9=34 位上数字之和（1） 3、（1）*3=（2） 34*3=102 4、从序号3开始求出奇数 8+6+4+2+0+6=34 位上数字之和（3） 5、（2）+（3）=（4） 102+34=128 6、用大于或等于结果（4）且为10最小整数倍的数 130-128=2 减去（4），其差即为所校验码X1=2-690123456789X113 1 3 12 11 10 9 8 7 6 5 4 3 2 1 6 9 0 1 2 3 4 5 6 7 8 9 2 X1, from the number two even get started 9 7 5 3 1 9 = 34 on the figures and (1), 3 (1)* 3 = (2) 34 3 = 102* 4, from the opening number 3 before sought odd 8 6 4 2 0 6 = 34 on the figures, and (3), 5 (2) (3) = (4) 102 34 = 128 6, with greater than or equal to the results (4) and as a 10 multiples of the minimum number 130-128 = 2 minus (4), the difference represents correcting code X1 = 2
Date : 2025-12-25 Size : 227kb User : 明明
[Other] 1121aircrack-2.41 DL : 0: 用于wfi无线密码的破解,在linux或WINDOWS下,最好不要用intel的无线网卡-wfi wireless password for the break in linux or windows, best not to use intel wireless card
Date : 2025-12-25 Size : 2.02mb User : 按时地方
[Other] ocIPRecord_V1.0 DL : 0: 1、Conn.asp 数据库连接文件 2、ipCheck.inc ip数据库查询函数 3、oc_ipRecord.asp ip记录文件（被调用文件） 4、oc_ipRecord.inc ip记录数据库 5、RecordIP.inc ip记录函数 6、showIPaddess.asp 显示IP来源页面 7、showIPRecord.asp 显示ip记录页面 8、Turn.inc 多用翻页函数 9、updateIPFrom.asp 更新IP来源数据库文件 10、worldip.asp 世界ip来源数据库 11、xg.css css文件 12、xmlHttpRequest.js XMLHttpRequest对象获取js文件 -1, Conn.asp database linking two documents, ipCheck.inc ip database query function 3, oc_ipRecord.asp ip records (to be called) 4. oc_ipRecord.inc 5 ip database records, RecordIP.inc 6 ip record function, showIPaddess.asp IP source revealed seven pages, Records show ip showIPRecord.asp eight pages, Turn.inc multi-function flip with nine, updateIPFrom.asp source IP update database files 10, worldip.asp world ip source database 11, 12 xg.css css document, xmlHttpRequest.js XMLHttpRequest object access documents js
Date : 2025-12-25 Size : 1.41mb User : 方杰
[Other] stackexp DL : 0: 控制台下的一个表达式计算器，支持带符号小数+-*/以及指数运算，可以输入（）来调整运算优先级，比如：输入1-2*(3+4^(1/2))=-9-Console expression under a calculator, a small number of support signed+-*/And index calculation, can enter () to adjust the priority operations, such as: Enter the 1-2* (3+ 4 ^ (1/2) ) =- 9
Date : 2025-12-25 Size : 125kb User : grin_t
[Other] source DL : 0: 编码转换的跨平台解决方案源代码名称：source 调试环境：Win XP/ Redhat 9；VC++ 6.0 / gcc3.2-Cross-platform data conversion solutions for the name of the source code: source debugging environment: Win XP/Redhat 9 VC++ 6.0/gcc3.2
Date : 2025-12-25 Size : 792kb User : 张建峰
[Other] ShuDu DL : 0: 每一行都用到1,2,3,4,5,6,7,8,9，位置不限，每一列都用到1,2,3,4,5,6,7,8,9，位置不限，每3×3的格子都用到1,2,3,4,5,6,7,8,9，位置不限，游戏的的过程就是用1,2,3,4,5,6,7,8,9填充空白，并要求满足每行、每列、每个九宫格都用到1,2,3,4,5,6,7,8,9-Each line used 1,2,3,4,5,6,7,8,9, location open, each column are used 1,2,3,4,5,6,7,8,9, Position Open, each 3 × 3 lattice are used 1,2,3,4,5,6,7,8,9, location open, the course of the game is to use 1,2,3,4,5 , 6,7,8,9 fill gaps and meet the requirements of each row, each row, each of squares are used 1,2,3,4,5,6,7,8,9
Date : 2025-12-25 Size : 99kb User : 黄
[Other] zgxq DL : 0: 本人机对弈程序采用了多种搜索算法.以下是本程序主要的类说明: 1.CEveluation类:估值类,对给定的棋盘进行估值. 2.CMoveGenerator类:走法产生器,对给定的棋盘局面搜索出所有可能的走法. 3.CSearchEngine类:搜索引擎基类. 4.CNegaMaxEngine类:负极大值法搜索引擎. 5.CAlphaBetaEngine类:采用了Alpha-Beta剪枝技术的搜索引擎. 6.CFAlphaBetaEngine类:fail-softalpha-beta搜索引擎. 7.CHistoryHeuristic类:历史启发类. 8.CAlphabeta_HHEngine类:带历史启发的Alpha-Beta搜索引擎. 9.CAspirationSearch类:渴望搜索引擎. 10.CIDAlphabetaEngine类:迭代深化搜索引擎. 11.CMTD_fEngine类:MTD(f)搜索引擎. 12.CTranspositionTable类:置换表. 13.CAlphaBeta_TTEngine类:加置换表的Alpha-Beta搜索引擎. 14.CPVS_Engine类:极小窗口搜索引擎. 15.CNegaScout_TT_HH类:使用了置换表和历史启发的NegaScout搜索引擎. 本程序具有悔棋,还原功能,可以记录走法,还可以进行布局.-err
Date : 2025-12-25 Size : 196kb User : david w
[Other] 3 DL : 0: 本程序实现计算任意长整数的加法运算，大致过程为显示提示信息-->用户输入—>显示结果。程序中字符集限定为’1’ ’2’ ‘3’ ‘4’ ‘5’ ‘6’ ‘7’ ‘8’ ’9’ ‘0’ ’,’ ‘ ’ ,长整数的长度不予限制。利用双向循环链表实现长整数的存储，每个结点含一个整型变量。任何整型变量的范围是-（215—1）~（215—1）。输入和输出形式：按中国对于长整数的表示习惯，每四位一组，组间用逗号隔开。-This procedure to achieve the calculation of arbitrary length integer addition operations, generally the process to show message-> User input-> result will be displayed. Program character set limited to 1 2 3 4 5 6 7 8 9 0, , a long integer that does not limit the length. The use of two-way cycle to achieve a long list integer storage, each node contains an integer variable. The scope of any integer variables is- (215-1) ~ (215-1). Input and output forms: according to China s long habit of integers that each of four groups of groups separated by commas.
Date : 2025-12-25 Size : 343kb User : 斜阳
[Other] vc DL : 0: 1) 具备十进制输入（0、1、2…8、9），扩展十六进制运算符（A、B…E、F），具备基本的运算功能，包括加法、减法、乘法、除法、取反，具备小数运算； 2) 三角函数运算，(反)正弦、(反)余弦、(反)正切、(反)余切； 3) 科学运算，包括乘方、取模、开方、指数、阶乘、对数、π和e，角度、弧度切换，二进制、八进制、十进制及十六进制数切换运算， 4) 具备历史计算的记忆功能 5) 用户操作非法或出错时，系统会显示提示； -1) with decimal input (0,1,2 ... 8,9), the expansion of hexadecimal operator (A, B ... E, F), equipped with the basic computing functions, including addition, subtraction, multiplication, division, check anti, with a small number of operations 2) trigonometric function computation, (Anti-) sinusoidal, (anti-) cosine, (anti-) tangent, (anti-) Cotangent 3) scientific computing, including involution, modulus, prescribing, index, factorial, logarithm, π and e, the angle of curvature of switching, binary, octal, decimal and hexadecimal number of switching operations, 4) with the history of computing and memory function 5) illegal or user error, the system will show the prompt
Date : 2025-12-25 Size : 2.7mb User : king
[Other] struts-1.2.9.jar DL : 0: struts-1.2.9.jar.zip,需要的话就用一下吧-struts-1.2.9.jar.zip, if necessary, by clicking on it
Date : 2025-12-25 Size : 496kb User : luoshihui
[Other] qishixunyou DL : 0: 编写程序求解骑士巡游问题：在n行n列的棋盘上（如n=5），假设一位骑士（按象棋中“马走日”的行走法）从初始坐标位置(x1，y1)出发，要遍访（巡游）棋盘中的每一个位置一次。请编一个程序，为骑士求解巡游“路线图”（或告诉骑士，从某位置出发时，无法遍访整个棋盘 — 问题无解）。当n=5时，意味着要在5行5列的棋盘的25个“点”处，按骑士行走规则，依次将1至25这25个“棋子”（数码）分别摆放到棋盘上（摆满25个位置则成功，否则失败问题无解）。例如，当n=5且初始坐标位置定为(1，1) — 即最左上角的那个点时，如下是一种巡游“路线图”。程序执行后的输出结果为： (x1,y1)? => (1=>5, 1=>5) : 1 1 1 6 15 10 21 14 9 20 5 16 19 2 7 22 11 8 13 24 17 4 25 18 3 12 23-Knight Parade programming to solve the problem: in n rows n columns of the board (such as n = 5), assuming a knight (in chess in the "horse to go on" walking method) from the initial coordinates (x1, y1) starting , to be sung (Parade) board in the position of each of the first. Please compile a program to solve for the Knight Parade "road map" (or tell the knight, starting from a position when not visiting with the board- the problem no solution).
Date : 2025-12-25 Size : 312kb User : 廖
[Other] Catch_That_Cow DL : 0: 上帝王牌 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute * Teleporting: FJ can move from any point X to the point 2 × X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it? Input Line 1: Two space-separated integers: N and K Output Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow. Sample Input 5 17 Sample Output 4 Hint The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.-shangdiwangpai Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X- 1 or X+ 1 in a single minute * Teleporting: FJ can move from any point X to the point 2 × X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it? Input Line 1: Two space-separated integers: N and K Output Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow. Sample Input 5 17 Sample Output 4 Hint The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
Date : 2025-12-25 Size : 176kb User : 上帝王牌
[Other] LCS DL : 0: 最长公共子序列问题描述与实验目的：序列Z=<B，C，D，B>是序列X=<A，B，C，B，D，A，B>的子序列，相应的递增下标序列为<2，3，5，7>。一般地，给定一个序列X=<x1,x2,…,xm>，则另一个序列Z=<z1,z2,…,zk>是X的子序列，是指存在一个严格递增的下标序列〈i1,i2,…,ik〉使得对于所有j=1,2,…,k使Z中第j个元素zj与X中第ij个元素相同。给定2个序列X和Y，当另一序列Z既是X的子序列又是Y的子序列时，称Z是序列X和Y的公共子序列。你的任务是：给定2个序列X、Y，求X和Y的最长公共子序列Z。输入样例 2 7 6 A B C B D A B B D C A B A 8 9 b a a b a b a b a b a b b a b b a 输出 Case 1 4 LCS(X,Y):B C B A Case 2 6 LCS(X,Y):a b a b a b -Longest common subsequence problem Description and Purpose: Z = <B,C,D,B> sequence is the sequence X = <A,B,C,B,D,A,B> the sequence, the corresponding sequence of increasing subscript <2,3,5,7 >. In general, given a sequence X = <x1,x2,…,xm>, then another sequence Z = <z1,z2,…,zk> is a sub-sequence of X, is the existence of a strictly increasing sequence subscript <i1,i2,…,ik> such that for all j = 1,2, ..., k so that the first j elements Z zj and X the same as the first element ij. Given two sequences X and Y, when another sequence Z as X, Y promoter sequence is a sub sequence, said sequence of X and Y Z is the common subsequence. Your task is: given two sequences X, Y, X and Y find the longest common subsequence Z. Input sample 2 76 A B C B D A B B D C A B A 89 b a a b a b a b a b a b b a b b a Output Case 1 4 LCS (X, Y): B C B A Case 2 6 LCS (X, Y): a b a b a b
Date : 2025-12-25 Size : 852kb User : wscnwps
[Other] 2 DL : 0: 对约束长度为9的viterbi硬判决进行测试-On the constraint length viterbi hard decision for the 9 test
Date : 2025-12-25 Size : 374kb User : ddd
[Other] remote-control DL : 0: 1. 远程屏幕查看及控制,录制远程屏幕到本地AVI文件及保存本地图片. 2. 远程摄象头管理,录制远程摄象头到本地AVI文件及监听远程语音. 3. 远程文件管理 4. 远程进程管理 5. 远程服务管理 6. 远程窗口管理 7. 远程命令行管理 8. 远程键盘记录 9. DDOS模块简单模拟 10. 远程3322动态域名上线测试-View and control remote screen, record remote screen to local AVI file and save the local picture. 2. Remote Xiangtou management, recording the remote Xiangtou local AVI file and monitor remote voice. 3. Remote file management. Remote process management. remote service management. remote window management. remote command-line management 8. remote keyloggers 9. DDOS module simple analog 10 remote 3322 dynamic domain name on the line test
Date : 2025-12-25 Size : 1.33mb User : ljfdf
[Other] CPPexam9 DL : 0: 1. 编写程序 Node.h 实现例 9-5 的节点类，并编写测试程序 lab9_1.cpp 实现链表的基本操作。 2. 编写程序 link.h 实现例 9-6 的链表类。在测试程序 lab9_2.cpp 中声明两个整型链表 A和 B，分别插入 5 个元素，然后把 B 中的元素加入 A 的尾部。 3. 编写程序 queue.h，用链表实现队列（或栈）类。在测试程序 lab9_3.cpp 中声明一个整型队列（或栈）对象，插入 5 个整数，压入队列（或栈），再依次取出并显示出来。 -1 Write a program to achieve cases 9-5 Node.h node classes and preparation of test procedures lab9_1.cpp realize list of basic operations. (2) Write a program link.h achieve cases 9-6 objectlist. In the test program lists lab9_2.cpp declare two integers A and B, five elements were inserted, and then B is added to A end of the elements. 3 programming queue.h, using chain queue (or stack) classes. In the test program lab9_3.cpp declare an integer queue (or stack) object, insert five integers, pressed into the queue (or stack), then turn out and displayed.
Date : 2025-12-25 Size : 635kb User : wolf
[Other] CPPexam9 DL : 0: 1. 声明 Point 类，有坐标_x，_y 两个成员变量；对 Point 类重载“++”（自增）、“--”（自减）运算符，实现对坐标值的改变。 2. 声明一个车（vehicle）基类，有 Run、Stop 等成员函数，由此派生出自行车（bicycle）类、汽车（motorcar）类，从 bicycle 和 motorcar 派生出摩托车（motorcycle）类，它们都有 Run、Stop 等成员函数。观察虚函数的作用。 1. 编写程序 Node.h 实现例 9-5 的节点类，并编写测试程序 lab9_1.cpp 实现链表的基本操作。 2. 编写程序 link.h 实现例 9-6 的链表类。在测试程序 lab9_2.cpp 中声明两个整型链表 A和 B，分别插入 5 个元素，然后把 B 中的元素加入 A 的尾部。 3. 编写程序 queue.h，用链表实现队列（或栈）类。在测试程序 lab9_3.cpp 中声明一个整型队列（或栈）对象，插入 5 个整数，压入队列（或栈），再依次取出并显示出来。 -1 Point class declaration, coordinates _x, _y two member variables pairs Point class overloads " ++" (increment), " -" (decrement) operator to achieve the coordinate values change. (2) declare a car (vehicle) base class, with Run, Stop and other member functions, and derives a bike (bicycle), automobiles (motorcar) class derived from the bicycle and motorcar motorcycle (motorcycle) class, which are with Run, Stop and other member functions. Observe the role of virtual function. 1 Write a program to achieve cases 9-5 Node.h node classes and preparation of test procedures lab9_1.cpp realize list of basic operations. (2) Write a program link.h achieve cases 9-6 objectlist. In the test program lists lab9_2.cpp declare two integers A and B, five elements were inserted, and then B is added to A end of the elements. 3 programming queue.h, using chain queue (or stack) classes. In the test program lab9_3.cpp declare an integer queue (or stack) object, insert fiv
Date : 2025-12-25 Size : 1007kb User : wolf
[Other] xuanzhuan DL : 0: 【问题描述】输入一个自然数Ｎ（2≤N≤9），要求输出如下的魔方阵，即边长为N*N，元素取值为1至N*N，1在左上角，呈顺时针方向依次放置各元素。 N=3时： 1 2 3 8 9 4 7 6 5 【输入形式】从标准输入读取一个整数N。【输出形式】将结果输出到文件文件file.out。输出符合要求的方阵，每个数字占5个字符宽度，向右对齐，在每一行末均输出一个回车符。【输入样例】 4 【输出样例】输出文件file.out内容为： 1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7-[Problem Description] Enter a natural number N (2 ≤ N ≤ 9), required output following magic square, that side of N* N, elements ranging from 1 to N* N, 1 in Top left, clockwise turn was placed each element. When N = 3: 123 894 765 [Input form] Read from the standard input one integer N. [Output] form The results output to file file.out. Output to meet the requirements of the square, each number representing five character width, aligned to the right, at the end of each line output a carriage return. [Sample input] 4 Sample output] [output file file.out says: 1234 1213145 1116156 10987
Date : 2025-12-25 Size : 372kb User : 服部半藏
[Other] tree DL : 0: (1) 由{4, 9, 0, 1, 8, 6, 3, 5, 2, 7}创建一棵二叉排序树bt并以括号表示法输出； (2) 判断bt是否为一棵二叉排序树； (3) 采用递归和非递归两种方法查找关键字为6的结点，并输出其查找路径； (4) 分别删除bt中的关键字为4和5的结点，并输出删除后的二叉排序树。 -(1) by {4, 9, 0, 1, 8, 6, 3, 5, 2, 7} to create a binary sort tree and bracket notation bt output (2) to determine whether a two bt Binary sort tree (3) the use of two methods of recursive and non-recursive lookup keyword node 6, and outputs its search path (4), respectively, to delete the keyword bt junction 4 and 5, and the output Binary sort tree after deletion.
Date : 2025-12-25 Size : 136kb User : Yoalnda
[Other] Day3 DL : 0: 题目：输入一个字符串，对于连续重复出现2-9次的字符，将重复的字符用次数加字符的方式转译，超过9次的用9次表示，只出现一次的字符要连续起来，并在两头用1封装起来，如果字符中包含1的话，用1来转译。 Eg: 1234444455 - 1112315425-char convert function
Date : 2025-12-25 Size : 832kb User : 余庆丹

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