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纯净的锁屏

于 2020-12-01 发布

 0  207

下载积分: 1 下载次数: 0

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代码说明：

说明：纯净的锁屏源，代码示例展示，可直接使用。(Pure lock screen source code demonstration)

文件列表：

LOCK.dsm, 4154330 , 2014-01-07
lock.ico, 894 , 2001-07-05
LOCK.identcache, 156 , 2010-09-11
LOCK.RES, 1004 , 2013-05-15
MAINLOCK.~ddp, 51 , 2013-05-15
MAINLOCK.~dfm, 674 , 2013-05-15
MAINLOCK.~PAS, 3848 , 2013-05-15
mainlock.dcu, 4303 , 2013-05-15
MAINLOCK.ddp, 51 , 2013-05-15
MAINLOCK.dfm, 599 , 2013-05-15
MAINLOCK.PAS, 3863 , 2013-05-15
md5.pas, 12579 , 2007-07-05
lklogin.~ddp, 51 , 2009-09-05
lklogin.~dfm, 155860 , 2009-09-05
lklogin.~pas, 2644 , 2007-11-16
lklogin.dcu, 6350 , 2013-05-15
lklogin.ddp, 51 , 2013-05-15
lklogin.dfm, 155860 , 2009-09-05
lklogin.pas, 2632 , 2013-05-15
LOCK.~DPR, 202 , 2013-05-15
LOCK.cfg, 434 , 2013-05-15
LOCK.dof, 2506 , 2013-05-15
LOCK.DPR, 190 , 2013-05-15
LOCK.dproj, 4903 , 2010-07-28
LOCK.dproj.local, 62 , 2010-07-28
LOCK.dsk, 3837 , 2014-01-07

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q1
说明： Pokeface 题目描述魔术师将一叠扑克顺次在桌上排成一行，初始全部正面朝上。随后，他每次挥一挥衣袖，都会翻转一连串的扑克，改变它们的正反朝向。从古代传下来的规矩，魔术师皆有师傅，他们要随时准备回答师傅的问题。问题形如：目前从第i张扑克牌到第j张扑克牌中，有几张正面朝上？作为魔术师的助手，你被要求学习程序设计、算法与数据结构。现在，魔术师把这个问题交给了你，你需要编写一个程序，维护相应的数据结构来保存桌子上牌面的情况，同时对于师傅的每一个问题，你要用尽量快的速度计算出答案。输入第一行包含三个正整数n、p和q。n表示扑克牌的数量，p表示魔术师的挥一挥次数，q表示师傅的问话次数。接下来共p+q行，每行包含一个操作，操作分为2种： H i j 魔术师挥一挥，改变了第i张牌到第j张牌的正反 Q i j 师傅问话：截至目前，第i张到第j张牌中有多少张正面朝上操作中恰好包含p次H操作，q次Q操作，与第一行的p和q吻合。H操作和Q操作的范围均包含边界（即第i张和第j张牌）。对于每个操作，输入保证1 ≤ i ≤ j ≤ n。输出对于每次Q操作，输出一行，包含一个整数，表示问题的答案。提示：线段树(Pokeface Title Description The magician will stack poker table sequentially in a row, all the initial face-up. He then every wisp will flip a series of poker pros and cons of changing their orientation. The rules handed down ancient times, the master magician Jie, they be ready to answer the teacher s questions. Question of the form: Currently playing cards i to j-card poker, there are a few face-up? As a magician s assistant, you are required to learn programming, algorithms and data structures. Now, the magician put this question to you, you need to write a program to maintain appropriate data structure to hold the cards on the table face the situation, while the master for each question, and you want to use as quickly calculate the answer. Enter The first line contains three positive integers n, p and q. n represents the number of poker, p represents the wave of the magic number, q represents the number of master s questioning. Next, a total of p+ q rows, each containi)

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积分：1
crt
this is a toolsthis is a toolsthis is a toolsthis is a tools(this is a tools)

2013-07-25 10:35:30下载

积分：1
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定义类IntChar，实现用一个整数保存一个长度不超过4的字符串。假设一个字符长度是1个字节。 * 例如："ABCD"，用一个整数保存，其二进制形式为：0100 0001 0100 0010 0100 0011 0100 0100 * 长度不足四个字符的串，剩下部分用0补足。 * * * 自己根据需要类IntChar的属性和方法，但需满足以下要求： * 1. 在类IntChar中，不能用C++语言中的string类型来定义任何属性（变量）。 * 2. 对于定义的类IntChar，在main函数中，能用以下方式使用类IntChar： * * void main() * { * IntChar IC("Love") * IC.ASC_Print() // 以字符串形式显示对象IC中保存的内容 * IC.Binary_Print() // 以二进制形式显示对象IC中保存的内容 * IC.Int_Print() // 以整数形式显示对象IC中保存的内容 * }(Custom class IntChar, save a length not exceeding 4 string with an integer. Suppose a character length is 1 byte.* For example: " ABCD" , kept by an integer, binary form: 0,100,000,101,000,010 0100 0011 0100 0100* length of less than four characters of the string, the rest of complement with 0. The*** the need class IntChar of the properties and methods, but must meet the following requirements:* 1. In class IntChar, can not use the string type the C++ language to define any attributes (variables).* 2. Defined class IntChar, the main function can be used following ways IntChar:** void main () {IntChar IC (" Love" ) * IC.ASC_Print () // in characters string displayed in the form saved in the object IC* IC.Binary_Print () // save the contents of the object IC* IC.Int_Print () in binary form // integer saved in the object IC*})

2013-05-11 22:10:57下载

积分：1
101259345hXeu7KjO
xzccxawefwefregref豆腐干的额外发懂得的 ( ccfdsgvergrtgreg)

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The "getSystemId()" method of a Notation node contains the system identifier associated with the notation, if one was specified.

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ROF系统接受端程序，和发送端程序进行相反的过程，恢复出原始的信号。(ROF system receiving end of the program, and the program sends the opposite end of the process, recover the original signal.)

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ProfilerProcessor
@fileoverview Profiler processor is used to process log file produced by V8 and produce an internal profile representation which is used for building profile views in "Profiles" tab.

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us_apgxy
zemax源码： This DLL models a standard ZEMAX surface type, either plane, sphere, or conic The surface also demonstrates a user-defined apodization filter The filter is defined as part of the real ray trace, case 5 The filter can be used at the stop to produce x-y Gaussian apodization similar to the Gaussian pupil apodization in ZEMAX but separate in x and y. The amplitude apodization is of the form EXP[-(Gx(x/R)^2 + Gy(y/R)^2)] The transmission is of the form EXP[-2(Gx(x/R)^2 + Gy(y/R)^2)] where x^2 + y^2 = r^2 R = semi-diameter The tranmitted intensity is maximum in the center. T is set to 0 if semi-diameter < 1e-10 to avoid division by zero. (ZEMAX source: This DLL models a standard ZEMAX surface type, either plane, sphere, or conic The surface also demonstrates a user-defined apodization filter The filter is defined as part of the real ray trace, case 5 The filter can be used at the stop to produce xy Gaussian apodization similar to the Gaussian pupil apodization in ZEMAX but separate in x and y. The amplitude apodization is of the form EXP [- (Gx (x/R) ^ 2+ Gy (y/R) ^ 2)] The transmission is of the form EXP [-2 (Gx (x/R) ^ 2+ Gy (y/R) ^ 2)] where x ^ 2+ y ^ 2 = r ^ 2 R = semi-diameter The tranmitted intensity is maximum in the center.T is set to 0 if semi-diameter <1e-10 to avoid division by zero.)

2006-11-10 17:25:16下载

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很简单的网络对时程序，命令提示符界面，基于windows网络通信与程序设计修改而成...
很简单的网络对时程序，命令提示符界面，基于windows网络通信与程序设计修改而成-Very simple network when procedures, command prompt interface, windows-based network communication and program design modifications from

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简单的VB文件分割、合并工具,有需要的就下载吧。(Simple VB file split, merge tool, there is a need to download it.)

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