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Description: 计算几何学常用算法库 C++语言实现
代码内容 计算几何学常用算法库，包括以下算法：
确定两条线段是否相交
判断点p是否在线段上
判断点q是否在多边形Polygon内
计算多边形的面积
寻找凸包 graham 扫描法
-geometry calculation commonly used algorithm for C language code as calculated geometry commonly used algorithm library, include the following algorithm : to determine whether the intersection of two segments to judge whether p-point of judgment, whether the point q Polygon Polygon estimate Polygon area for convex hull graham scanning
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Size: 12980 |
Author: henry |
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Description: 计算几何学常用算法库 C++语言实现
代码内容 计算几何学常用算法库，包括以下算法：
确定两条线段是否相交
判断点p是否在线段上
判断点q是否在多边形Polygon内
计算多边形的面积
寻找凸包 graham 扫描法
-geometry calculation commonly used algorithm for C language code as calculated geometry commonly used algorithm library, include the following algorithm : to determine whether the intersection of two segments to judge whether p-point of judgment, whether the point q Polygon Polygon estimate Polygon area for convex hull graham scanning
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Size: 12288 |
Author: henry |
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Description: acm.pku.edu.cn/JudgeOnline上的题目1001到1010的十道题的代码，都是已经通过的，贴到.txt文件中的。-acm.pku.edu.cn/JudgeOnline on the topic of the 1001-1010 title 10 of the code, are already adopted and put into. txt files.
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Size: 7168 |
Author: captain |
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Description: 实现lzw词典编码方法
Lzw编码算法的设计思路：
1.开始时的词典包含所有的根（先将所有单个字符编码），当前缀P时空的；
2.当前字符（C）=字符流中的下一个字符；
3.判断缀-符串P+C是否在词典中
A如果“是”：P=P+C//(用C扩展P)
B．如果“否”：
a 把代表当前前缀P的马子输出到码字流；
b 把缀-符串P+C添加到词典；
c 令P=C//（现在的P仅包含一个字符C）；
4. 判断字符流中是否还有码字要译
（1）如果是，就返回到2；
（2）如果否
a 把代表当前前缀P的码字输出到码字流；
b 结束。.
-Coding method to achieve lzw dictionary coding algorithm Lzw design ideas: 1. The beginning of the dictionary contains all the root (first of all a single character encoding), the current time is made up of P 2. The current character (C) = the next character stream character 3. judge up- at string P+ C is in the dictionary, if A " is" : P = P+ C// (use C extension of P) B. If " no" : a prefix P to the representative of the current output of the commode to the code word stream b put up- at string P+ C added to the dictionary c so that P = C// (the P now contains only one character C) 4. to determine whether there are character stream code words to translate (1) If yes, on the return to the 2 (2) If it has a prefix P to the representative of the current output of the code word to code word stream b end. .
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Size: 1024 |
Author: 符晓娟 |
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Description: 输入一个一阶语言，开头和结尾必须是括号。例如，(@a(P(a)).
数理逻辑的应用：通过C语言进行一阶语言的判断约束符号只限P,Q,R 变员符号只限a,b,c 命题符号为:&,|,~,-(蕴含),=(等价)
量词符号:全称量词:@ 存在量词:# -Enter a first order language must start and end brackets. For example, (@ a (P (a)). The application of mathematical logic: first-order through the C language sign language to judge bound only P, Q, R become members of symbols restricted to a, b, c proposition symbols :& ,|,~,-( implication), = (equal) quantifier symbols: universal quantifier: @ existential quantifier:#
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Size: 2048 |
Author: 涵 |
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Description: 验证代理级别的PHP代码
去掉Referer的判断，避免代理通过统计屏蔽代理验证通过
增加读取IP的方式
增加对代理国家的判断
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Size: 2048 |
Author: Adam Hamilton |
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Description: Uva Judge Online Problem 101 3n+1 Really easy but too interesting
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Size: 109568 |
Author: Roger |
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Description: /*题意：判断两幅图是否同构 图中即判断图中环和链的个数是否相同 环和链是否一一相对应
利用并查集 一个集合中若人数num[]和拉手对手p[]相等 则为环，num[]==p[]+1则为链*/-/* The meaning of problems: to determine whether the two images with a composition that judge whether each of the corresponding figure rings and chains are the same number of rings and chain
A collection if the number of utilization and check set num [] and handle opponents p [] equal compared to ring, num [] == p []+1 compared to chain* /
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Size: 1024 |
Author: |
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Description: Uva Judge Online Problem 101 3n+1 Really easy but too interesting -Uva Judge Online Problem 101 3n+1 Really easy but too interesting
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Size: 109568 |
Author: 原北京 |
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Description: 参数：
p1～p4：直线上不相同的两点
*p：通过指针返回结果
返回值：1：两直线相交；2：两直线平行
如需要判断两线段交点，检验k 和对应k 1（注释中）的值是否在0～1 之间，用在0～1
之间的那个求交点
-Parameters:
p1～p4：直线上不相同的两点
P1 ~ P4: two points in a straight line.
*p：通过指针返回结果
*p: the result is returned by the pointer.
返回值：1：两直线相交；2：两直线平行
Return value: 1: two straight line two: 2
如需要判断两线段交点，检验k 和对应k 1（注释中）的值是否在0～1 之间，用在0～1
If you need to judge the intersection of two line segments, test K and the value of the corresponding K 1 (notes) is between 1 to 0, with the 0 to 1
之间的那个求交点
Between the intersection
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Size: 17408 |
Author: machang |
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Description: 1.需求分析
本题要求循环数数，数到某一个数之后将其排除并从下一个开始重新开始数数，直到剩下一个人为止，要求剩下的人标号为1。我组有两份代码，第一个程序中，输入n值，并数的数至多为3p（p为前n个数最小公倍数，3p为了能够找到规律，n>=7时由于最小周期过大在下面的表格中不便列举），将每种可能的x,y值分组罗列出来，并试图从中找出规律，一旦规律出现，那么我们可以随意举一个符合规律的其他例子来进行验证（验证代码为另一份，而且验证的时候y不再有范围限制，只要符合规律即可）
2.概要设计
本题设计一个循环链表，用来存储n名队员的编号和存活情况（当然了，如果执行任务后安全返回就不需要继续数数了，因此执行任务即视为牺牲），两份代码中均含有两个自定义函数：
soldier *create(int n)//用于创建一个长度为n的循环链表
int judge(soldier *&head,int n)//用于观察小队中人数是否为1
第一份还含有以下两个函数：
int f(int a,int b)//为下面函数服务
int mf(int n)//用来计算前n个数最小公倍数
这两个函数用来帮助我们寻找规律
-1. After the needs analysis of this question requires loop count, count to a certain number of their exclusion and a re-start to count the start until only one person so far, the requirements for the remainder of the label 1. I have two groups of code, the first program, enter the value of n, the number and the number is up to 3p (p n is the number before the least common multiple, 3p in order to find the law, n> = 7 Because the minimum period is too large in the following table lists the inconvenience), the every possible x, y values grouping set out and tried to find out the rules, once the law is present, we are free to cite other examples of a line with the law to carry out verification (authentication code to another, and to verify the time y no limitation, so long to comply with the law) 2. outline of the design problem to design a circular linked list for storing n number of players and survival (of course, if you perform a task after the safe return of a few do not need to
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Size: 1024 |
Author: alex |
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Description: authorware7.0 作业
1. 使用authorware7.0工具，设计一段自拟主题的动画，立意要新颖。
3. 作品中要至少使用一次交互图标，一次判断图标、一次移动图标。
4. 作品中至少应添加一幅GIF动画(运行过程中按下ctrl+p可选中该GIF，然后可以改变大小)
5. 作品要添加背景音乐。
6. 作品名称要与主题相符，内容有创意。
-Authorware7.0 job
1 the use of authorware7.0 tools, design a theme with animation, willing to novel.
3 works to use at least once interactive icon, a judge icon, once moving icon.
4 works should add at least a GIF animation (run the process press ctrl+p can the GIF, and then can change the size)
5 works to add background music.
6 the name of the work should be consistent with the theme, content creative.
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Size: 28646400 |
Author: h |
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