E - Rails Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extrem…

思路:将出车站的顺序存入数组train,由于入车站的顺序是固定的,为1~N,所以用P表示进站的车,初始为1. 接下来举例说明吧: 原来入站顺序:    1 2 3 4 5 读入的出战顺序: 3 4 2 5 1 按照train数组的顺序来执行, 1.一开始p=1,i=1: p与train[i]=3不相等,将p(1)入栈,p++:再比较不相等,将p(2)入栈,p++: p=train[3],则i++,p++: 2.i=2: 先比较train[i]与栈顶元素2是否相同,不相同,则与p比较. train…

  Rails  There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover,…

题意:一列有n节车厢的火车按顺序进站,给你一个出站顺序,问你该火车的车厢能否以该顺序出站? 分析:出站的车厢满足后进先出的关系,所以我们考虑采用栈s 假设车厢一共有n节,n = 5: 进站顺序A:1 2 3 4 5 出站顺序B/target: 2 3 4 5 1 如果进站的车厢没有立马出站,则将该车厢入栈s.push(A++) 判断栈里面车厢的顺序是否和出站顺序一致 s.top()==  target[B] #include<cstdio> #include<stack> usin…

本章,你将扩大你的模型测试,测试整个Rails栈的逻辑(从请求到回复,使用端到端测试). 使用Capybara来帮助写end-to-end 测试. 好的测试风格,包括端到端测试,大量目标明确的单元测试,和相关的一些覆盖中间代码的测试. 开始写Rails Requirements-gathering,分析需求,是一整本书的内容.本节假设是写一个自用的小程序,因此无需military-grade precision. 列出非正式的需求单子: A user can enter a task, asso…

问题如下: 问题 B: Rails 时间限制: Sec 内存限制: MB 提交: 解决: [提交][状态][讨论版] 题目描述 There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was…

 Rails  There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, i…

从 rails 窥探 web 全栈开发(零) 本文将讲述在学习之前几个必须要知道的概念,这些词汇在 rails 中都会出现. 本文前置条件:安装好 Ruby. 从 rails 窥探 web 全栈开发(零) 先换源 Ruby RVM Rails RubyGems Gem Gemfile Bundle Rake Rakefile 先换源 gem sources -l gem sources --add https://gems.ruby-china.com/ --remove https://rub…

题目链接: https://cn.vjudge.net/problem/UVA-514 /* 问题 输入猜测出栈顺序,如果可能输出Yes,否则输出No 解题思路 貌似没有直接可以判定的方法,紫书上给出的方法是在进行一个入栈出栈操作,看是否能够构成原始的入栈序列. */ #include<cstdio> #include<stack> using namespace std; int ok(int a[],int n); int main() { freopen("E:\\…

题意:判断出栈顺序是否合法 题解:两个指针,A指向入栈序列,B指向出栈. 的分三种情况:if     1.A==B :直接入栈加出栈即可A++,B++ else 2.和栈顶相同,直接出栈A==stack.top   A++,stack.pop else 3. 都不符合 压入栈stack.push(A) #define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<cstring> #include<queue>…