sign

结对作业2

坚强是说给别人听的谎言 提交于 2020-02-27 08:21:06
软件工程结对作业2 小学生计算题系列 题目1: 我们在刚开始上课的时候介绍过一个小学四则运算自动生成程序的例子,请实现它,要求: 能够自动生成四则运算练习题 可以定制题目数量 用户可以选择运算符 用户设置最大数(如十以内、百以内等) 用户选择是否有括号、是否有小数 用户选择输出方式(如输出到文件、打印机等) 最好能提供图形用户界面(根据自己能力选做,以完成上述功能为主) #include <ctime> #include<iostream> #include<stdlib.h> #include <cstdlib> #include <sstream> #include<stdio.h> #include<string.h> #include <fstream> using namespace std; #define random(a,b)(rand()%(b-a+1)+a) const string filename="test000.txt"; char yunsuanfu(int a) { switch(a) { case 1:return '+' ; break; case 2:return '-' ; break; case 3:return '*' ; break; case 4:return '/' ; break; } } int main() { const

Java程序设计之扑克牌

末鹿安然 提交于 2020-02-25 19:54:55
  这段代码的主要实现功能扑克牌的洗牌和发牌功能,一副牌,红桃,黑桃,梅花,方片,A~K,不含大小王。   构造一个class。   首先是声明花色: private String[] sign={"方片","红桃","黑桃","梅花"};//扑克的四种花色   然后是洗牌功能,意味花色与A~K结合构造成顺序的52张扑克牌,构造一个ArrayList对象存储扑克牌的值。 private static ArrayList<String> List;   接下来再类中实现此方法,构造出一副扑克牌。 public void Init(){ List = new ArrayList(); for(String str:sign){ for(int i = 1;i<=13;i++){ String extra_sign = ""; //将数字与扑克匹配 if(i == 1){ extra_sign = "A"; List.add(str+extra_sign); }else if(i == 11){ extra_sign = "J"; List.add(str+extra_sign); }else if(i == 12){ extra_sign = "Q"; List.add(str+extra_sign); }else if(i == 13){ extra_sign = "K"; List

没有路线匹配“ / users / sign_out”设计导轨3

不羁的心 提交于 2020-02-25 15:26:25
我已经在我的应用程序上安装了devise,并在我的 application.html.erb 文件中应用了以下内容: <div id="user_nav"> <% if user_signed_in? %> Signed in as <%= current_user.email %>. This cannot be cheese? <%= link_to 'Sign out', destroy_user_session_path %> <% else %> <%= link_to 'Register', new_user_registration_path %> or <%= link_to 'Sign in', new_user_session_path %> <% end %> </div> 我跑了 rake routes 并确认所有路线均有效。 另外,在我的 routes.rb 文件中,我有 devise_for :users 和 root :to => "home#index" 。 单击“注销”链接时,出现以下路由错误: No route matches "/users/sign_out" 任何想法导致错误的原因是什么? #1楼 在“ devise_for:users”之前查看您的routes.rb是否具有“ resource:users”,然后尝试交换它们: 作品

jmeter读取本地CSV文件

本小妞迷上赌 提交于 2020-02-25 07:26:14
用jmeter录制考试上传成绩等脚本时,出现的问题及解决方法如下: 1、beanshell前置处理器,不能读取本地csv文件里的数据: 方法一: 在beanshell里不能直接从本地的csv文件里读取到tid的值,幸好tid数据是有规律的,从700000开始,依次增加,于是通过如下的几个步骤,解决了我的问题: (1)首先在test plan增加了一个全局变量,如tid,并赋值700000(这时,就不能读取csv里的数据了,这里需要注意); (2)在beanshell前置处理器上写如下代码: 这里的md5加密是导入的jar包,进行调用的 import com.xuexin.*; import java.security.MessageDigest; import java.security.NoSuchAlgorithmException; String tid=vars.get("tid"); String m_xuexin_id = Integer.toString(Integer.parseInt(tid)+1); String examSubjectID = "******"; String mid_url = "examSubjectID=" + examSubjectID + "&xuexinID=" + m_xuexin_id; mymd5 mymd5 = new

Codeforces C. NP-Hard Problem 搜索

醉酒当歌 提交于 2020-02-25 00:31:46
C. NP-Hard Problem time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting. Suppose the graph G is given. Subset A of its vertices is called a vertex cover of this graph, if for each edge uv there is at least one endpoint of it in this set, i.e. or (or both). Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of

PTA 甲级1006 Sign In and Sign Out

跟風遠走 提交于 2020-02-22 23:44:28
目录 PTA 甲级 1006 Sign In and Sign Out (25 分) Input Specification: Output Specification: Sample Input: Sample Output: 生词 关键句子 注意点 PTA 甲级 1006 Sign In and Sign Out (25 分) 题目原文 At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked and locked the door on that day. Input Specification: Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M

【Vue常用指令】

天涯浪子 提交于 2020-02-20 12:31:49
目录 v-html v-text v-for v-if v-show v-bind v-on v-model 指令修饰符 计算与侦听属性 自定义属性 获取DOM元素 "@ *** Vue.js官方给自己的定义为==数据模版引擎==,并给出了一套渲染数据的指令。本文将详细介绍Vue.js的常用指令 导入vue.js https://cdnjs.cloudflare.com/ajax/libs/vue/2.5.16/vue.min.js Vue.js使用了基于HTML的模版语法,使用vue最简单的方式就是渲染数据,渲染数据最常见的形式就是使用"Mustache"语法(双大括号)的文本插值。 ==- - - - - - - - - - - - - - - - - - - - - -== 一个简单的示例: <body> <div id="sign">{{ message }}</div> <script> let sign = new Vue({ el: '#sign', data: { message: 'Hello Vue!', }, }); </script> </body> 首先,创建一个vue实例,并在创建实例的过程中传入一个对象。 · 该对象的第一个属性名为el,它的值是我们需要渲染的目标标签,我们通过属性查找定位这个标签。 · 该对象的第二个属性名为data

A1006 Sign In and Sign Out (25分)

◇◆丶佛笑我妖孽 提交于 2020-02-15 03:45:49
At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked and locked the door on that day. Input Specification: Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format: ID_number Sign_in_time Sign_out_time where times are given in the format HH:MM:SS, and ID

iOS - 如何判断一个点是不是在方框(CGRect)、圆(Circle)、三角形(Triangle)内呢?

笑着哭i 提交于 2020-02-13 23:21:11
如何判断一个点是不是在方框(CGRect)、圆(Circle)、三角形(Triangle)内呢? 1. 方框 //苹果官方方法可以判断 + ( BOOL ) point : ( CGPoint ) point inSquareArea : ( CGRect ) rect { return CGRectContainsPoint ( rect , point ) ; } 2. 圆 //圆心到点的距离>?半径 + ( BOOL ) point : ( CGPoint ) point inCircleRect : ( CGRect ) rect { CGFloat radius = rect . size . width / 2.0 ; CGPoint center = CGPointMake ( rect . origin . x + radius , rect . origin . y + radius ) ; double dx = fabs ( point . x - center . x ) ; double dy = fabs ( point . y - center . y ) ; double dis = hypot ( dx , dy ) ; return dis <= radius ; } 3. 三角形 //点都否在三边线的右边?这个应该不是最优解 + ( BOOL

PAT(Advanced)甲级---1006 Sign In and Sign Out (25 分)

不羁岁月 提交于 2020-02-12 22:45:53
At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked and locked the door on that day. Input Specification: Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format: ID_number Sign_in_time Sign_out_time where times are given in the format HH:MM:SS, and ID