private void toolStripButton32_Click(object sender, EventArgs e) { /重叠部分去除操作——测试成功 if (mapMain.Layers.Count == 0) { return; } //重叠分析 //遍历要素,显示面积 FeatureSet fs = null; fs = Lzq_LayerManager.getFeatureSetByName(layerNamePolygon, mapMain); if (fs.Featur

求点到直线的距离: double dis(point p1,point p2){   if(fabs(p1.x-p2.x)<exp)//相等的  {    return fabs(p2.x-pegx);    }  else     {   double k=(p2.y-p1.y)/(p2.x-p1.x);   double b=p2.y-k*p2.x;   return fabs(k*pegx-pegy+b)/sqrt(k*k+1);//返回的是距离的   }}判断多边形是否为凸多边形 if

Shape of HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4972    Accepted Submission(s): 2250 Problem Description 话说上回讲到海东集团推选老总的事情,最终的结果是XHD以微弱优势当选,从此以后,“徐队”的称呼逐渐被“徐总”所取代,海东集团(HDU)也算是名副其实了.

链接: http://acm.hdu.edu.cn/showproblem.php?pid=2108 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27459#problem/A Shape of HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4320    Acc

1.1 环境以及变量的定义.赋值.展开.删除 export:将一个变量导入到环境中:export PATH=$PATH:/home. readonly 讲一个变量设置为只读模式,在shell脚本中定义字面值常量特别有用:readonly days_per_week=7, readonly -p 打印当前环境只读模式变量. env可以对命令执行时的环境做更细致的控制:env -i PATH=$PATH HOME=$HOME awk '...' file1, 执行awk命令时的环境仅仅为env指定的

大致思路:首先对于所给的洞的点,判断是否是凸多边形,图形的输入和输出可以是顺时针或者逆时针,而且允许多点共线 Debug 了好几个小时,发现如下问题 判断三点是否共线,可用斜率公式判断 POINT point_A, point_B, point_C; if(point_A.x == point_B.x || point_B.x == point_C.x){ if(point_A.x == point_B.x && point_B.x == point_C.x) continue; }els

Problem Description 话说上回讲到海东集团推选老总的事情,最终的结果是XHD以微弱优势当选,从此以后,"徐队"的称呼逐渐被"徐总"所取代,海东集团(HDU)也算是名副其实了.创业是需要地盘的,HDU向钱江肉丝高新技术开发区申请一块用地,很快得到了批复,据说这是因为他们公司研发的"海东牌"老鼠药科技含量很高,预期将占全球一半以上的市场.政府划拨的这块用地是一个多边形,为了描述它,我们用逆时针方向的顶点序列来表示,我们很想了解这块地

//使用for循环 判断是否有重名 var len=$("li").length;//获取页面中所有li的数量 for(var i=0; i<len; i++){ oldname_wenjiaming=$("li:eq("+i+") #wenjianming").val(); //循环获得每个li下面的#wenjianming的值 if(checkname==oldname_wenjiaming){ alert("文件名有重复!&

一.判断单个列表中的元素是否存在重复 使用set方法去重后,和原list进行对比,如果相等,那么说明原列表无重复,如果存在重复,说明列表存在重复 def is_repect_all(L): repeatList = []; setList = set(L); flag=True; if len(L) != len(setList): flag=False; print('列表的id存在重复,其中重复项及重复次数如下:'); for each_item in setList: re_count =

问题 H: Hunter's Apprentice 时间限制: 1 Sec  内存限制: 128 MB 提交: 353  解决: 39 [提交] [状态] [命题人:admin] 题目描述 When you were five years old, you watched in horror as a spiked devil murdered your parents. You would have died too, except you were saved by Rose, a pass

描述 While creating a customer logo, ACM uses graphical utilities to draw a picture that can later be cut into special fluorescent materials. To ensure proper processing, the shapes in the picture cannot intersect. However, some logos contain such inte

Hunter's Apprentice 题目描述 When you were five years old, you watched in horror as a spiked devil murdered your parents. You would have died too, except you were saved by Rose, a passing demon hunter. She ended up adopting you and training you as her ap

public static int repeatedStringMatch(String A, String B) { //判断字符串a重复几次可以包含另外一个字符串b,就是不断叠加字符串a直到长度大于等于b,叠加一次计数+1 //看现在的字符串是否包含b,包含就返回,不会包含就再叠加一次,因为可能有半截的在后边,再判断,再没有就返回-1 int count = 0; StringBuilder sb = new StringBuilder(); while (sb.length() < B.l

题意不难理解,给出多个多边形,输出多边形间的相交情况(嵌套不算相交),思路也很容易想到.枚举每一个图形再枚举每一条边 恶心在输入输出,不过还好有sscanf(),不懂可以查看cplusplus网站 根据正方形对角的两顶点求另外两个顶点公式: x2 = (x1+x3-y3+y1)/2; y2 = (x3-x1+y1+y3)/2; x4= (x1+x3+y3-y1)/2; y4 = (-x3+x1+y1+y3)/2; 还有很多细节要处理 #include <iostream> #include &

链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=10 Area Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Jerry, a middle school student, addicts himself to mathematical research. Maybe the problems he has thought are

知识点: .sort()方法用于对数组元素排序,并返回数组. var _arr = ['旅行箱', '旅行箱', '小米', '大米']; var _res = []; // _arr.sort(); for (var i = 0; i < _arr.length;) { var count = 0; for (var j = i; j < _arr.length; j++) { if (_arr[i] == _arr[j]) { count++; } } _res.push([_arr[i]

Set<User> userSet = new HashSet<>(); User user1= new User("aa","11"); User user2= new User("aa","11"); userSet.add(user1); userSet.add(user2); System.out.println("set size"+userSet.size()); Ite

之前碰到asp.net core异步进行新增操作并且需要判断某些字段是否重复的问题,进行插入操作的话会导致数据库中插入重复的字段!下面把我的解决方法记录一下,如果对您有所帮助,欢迎拍砖! 场景:EFCore操作MySql数据库的项目,进行高并发插入操作 需求:消息队列,最后进行新增数据的操作,插入前判断某些字段是否重复 问题:采用await db.SaveChangesAsync()进行提交操作前,FirstOrDefault判断数据库中是否有重复数据.测试100条一样的数据进行并发插入,结果数

题目: 给个n个点的多边形,n个点按顺序给出,给个点m,判断m在不在多边形内部 题解: 网上有两种方法,这里写一种:射线法 大体的思想是:以这个点为端点,做一条平行与x轴的射线(代码中射线指向x轴正方向) 如果交点个数为奇数的话就在内部,如果为偶数(包括0)就在外部 #include<cstdio> #include<algorithm> #include<cstring> #define N 105 using namespace std; int n,m; stru

C#实现多级子目录Zip压缩解压实例 参考 https://blog.csdn.net/lki_suidongdong/article/details/20942977 重点: 实现多级子目录的压缩,类似winrar,可以选择是否排除基准目录 1 public void ZipDirectoryTest() 2 { 3 string path = System.IO.Path.Combine(System.IO.Path.GetTempPath(), DateTime.Now.Ticks.ToS

-------下面使用标量值函数判断  出现重复的个数 create function fn_str_times(@str varchar(1000),--原子符串@indexstr varchar(20)--查找的字符)returns intasbegindeclare @findlen intdeclare @times intset @findlen=LEN(@indexstr)set @times=0while(charindex(@indexstr,@str))>0BEGINset @