苏州其然软件开发培训

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韩奇峰高级讲师

多年实战工作经验曾参与制作宝马Usage Training项目、DMS项目，奥迪全 息投影项目，奔驰等多家汽车门户行业大型项目，负责UI设计、界面设计、3D模型制作、前端开发等职务。

从事设计行业多年，精通PhotoShop、UI设计、AfterEffects、Flash、 Actionscript、HTML、CSS、JavaScript、jQuery、资深动画设计师，设计作品曾获得全国动画设计三等奖。

课程讲解注重实战应用，对讲述知识点穿插案例制作，使课程内容更加接近 工作中实际的项目。授课风格注重实战经验分析，深受学生喜欢。

Java工程师的工资待遇怎么样?

Java工程师的工资待遇怎么样?

Java软件工程师一般月薪范围在4000-10000元，远远超过了应届毕业生月薪 2500元的平均水平。通常来说，有一年工作经验的Java高级软件工程师的薪酬大致在年薪10—13万左右。

从Java的应用领域来分，Java语言的应用方向主要表现在以下三个方面：首 先是大中型的商业应用;其次是桌面应用，就是常说的C/S应用;再次是移动领域应用。

综上而言JAVA就业方向为：可以从事JSP网站开发、Java编程、Java游戏开 发、Java桌面程序设计，以及其他与Java语言编程相关的工作。可进入电信、银行、保险专业软件开发公司等从事软件设计和开发工作。

JAVA 分布式大纲

一阶段 java基础，我们将学习变量，基本数据类型，进制，转义字符，运 算符，分支语句和循环语句等，以达到训练基础语法和逻辑能力的目的。还有对数组、面向对象和异常处理等。

二阶段 javaWeb，主要是学习Web前端开发基础和框架、Servlet和JSP在Web 后端的应用、Web后端开发相关专题、MVC和分层架构以及项目开发流程及CASE工具的使用等。

三阶段 java框架，像框架整合开发(SSH/SSS)、RESTful架构和移动端接口 设计、第三方接口和在线支付功能、网站安全和Spring Security应用实战、复杂用户交互处理和Spring Web Flow的应用、MyBatis的应用和SSM整合等 技术点都是需要你掌握的。

四阶段 java 云数据，亿级并发架构演进、Linux基础、搭建tomcat环境以 及大数据开发云计算等高级Java教程，是Java技术的高端知识。其中穿插项目实战演练，企业真实项目供学员应用学习，进行知识体系的“二次学习” 。

hdu1077【计算几何，枚举圆心】

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Catching Fish

Time Limit: 10000/5000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1930 Accepted Submission(s): 784

PRoblem Description Ignatius likes catching fish very much. He has a fishnet whose shape is a circle of radius one. Now he is about to use his fishnet to catch fish. All the fish are in the lake, and we assume all the fish will not move when Ignatius catching them. Now Ignatius wants to know how many fish he can catch by using his fishnet once. We assume that the fish can be regard as a point. So now the problem is how many points can be enclosed by a circle of radius one.

Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.

Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case starts with a positive integer N(1<=N<=300) which indicate the number of fish in the lake. Then N lines follow. Each line contains two floating-point number X and Y (0.0<=X,Y<=10.0). You may assume no two fish will at the same point, and no two fish are closer than 0.0001, no two fish in a test case are approximately at a distance of 2.0. In other Words, if the distance between the fish and the centre of the fishnet is smaller 1.0001, we say the fish is also caught.

Output For each test case, you should output the maximum number of fish Ignatius can catch by using his fishnet once.

Sample Input 4 3 6.47634 7.69628 5.16828 4.79915 6.69533 6.20378 6 7.15296 4.08328 6.50827 2.69466 5.91219 3.86661 5.29853 4.16097 6.10838 3.46039 6.34060 2.41599 8 7.90650 4.01746 4.10998 4.18354 4.67289 4.01887 6.33885 4.28388 4.98106 3.82728 5.12379 5.16473 7.84664 4.67693 4.02776 3.87990 20 6.65128 5.47490 6.42743 6.26189 6.35864 4.61611 6.59020 4.54228 4.43967 5.70059 4.38226 5.70536 5.50755 6.18163 7.41971 6.13668 6.71936 3.04496 5.61832 4.23857 5.99424 4.29328 5.60961 4.32998 6.82242 5.79683 5.44693 3.82724 6.70906 3.65736 7.89087 5.68000 6.23300 4.59530 5.92401 4.92329 6.24168 3.81389 6.22671 3.62210

Sample Output 2 5 5 11

题解： 依次把圆心枚举出来，然后取**大值 【盗图】原文链接

代码：

#include <iostream> #include <string> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <algorithm> #include <queue> #include <map> #include <sstream> #define Maxn 100005 using namespace std; double pos[305][2]; double centre1[2], centre2[2]; int n; void GetCentre(double A[], double B[]) { double x0 = (A[0] B[0]) / 2; double y0 = (A[1] B[1]) / 2; double a = sqrt( 1 - ( (A[0] - B[0]) * (A[0] - B[0]) (A[1] - B[1]) * (A[1] - B[1]) ) / 4); if (fabs(A[1] - B[1]) < 1e-6) { centre1[0] = centre2[0] = x0; centre1[1] = y0 a; centre2[1] = y0 - a; } else { double angel = atan((B[0] - A[0]) / (A[1] - B[1])); centre1[0] = x0 a * cos(angel); centre1[1] = y0 a * sin(angel); centre2[0] = x0 - a * cos(angel); centre2[1] = y0 - a * sin(angel); } } int CatchFish() { int ans1 = 0; int ans2 = 0; for (int i = 0; i < n; i ) { if (sqrt( (pos[i][0] - centre1[0]) * (pos[i][0] - centre1[0]) (pos[i][1] - centre1[1]) * (pos[i][1] - centre1[1]) ) < 1.0001) ans1 ; if (sqrt( (pos[i][0] - centre2[0]) * (pos[i][0] - centre2[0]) (pos[i][1] - centre2[1]) * (pos[i][1] - centre2[1]) ) < 1.0001) ans2 ; } return max(ans1, ans2); } int main() { int T, i, j; cin >> T; while (T--) { int ans = 1; // 注意n为1时 scanf("%d", &n); for ( i = 0; i < n; i ) { scanf("%lf %lf", &pos[i][0], &pos[i][1]); } for ( i = 0; i < n; i ) for ( j = i 1; j < n; j ) { // 如果两点距离超过2，直接略过，不然会超时 if ((pos[i][0] - pos[j][0]) * (pos[i][0] - pos[j][0]) (pos[i][1] - pos[j][1]) * (pos[i][1] - pos[j][1]) > 4.0) continue; GetCentre(pos[i], pos[j]); ans = max(ans, CatchFish()); } printf("%d\n", ans ); } }

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