Rails_via牛客网
题目
链接:https://ac.nowcoder.com/acm/contest/28537/D
来源:牛客网时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld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 possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, …, N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, …, aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.
Input
The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, …, N. The last line of the block contains just 0.The last block consists of just one line containing 0.
输入描述:
The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, ..., N. The last line of the block contains just 0.
The last block consists of just one line containing 0.
输出描述:
The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null'' block of the input.
示例1
输入
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
输出
Yes
No Yes
题解
这道题目其实不难,重点是输入以及输出有大大的问题
代码
#include <iostream>
#include <stack>
using namespace std;
int s[1005];
int n = 0;
bool check()
{
int j = 0;
stack<int> a;
for (int i = 1; i <= n; i++)
{
a.push(i);
while (!a.empty() && a.top() == s[j])
{
a.pop();
j++;
}
}
if (a.empty())
{
return true;
}
else
return false;
}
int main()
{
int flags = 0;
int buf = 0;
while (1)
{
cin >> buf;
if (!buf)
break;
else
{
n = buf;
if (flags)
cout << endl;
}
while (1)
{
cin >> buf;
if (buf)
{
s[0] = buf;
for (int i = 1; i < n; i++)
cin >> s[i];
flags = 1;
if (check())
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else
break;
}
}
return 0;
}
Rails_via牛客网的更多相关文章
- 牛客网 --java问答题
http://www.nowcoder.com/ 主要是自己什么都不怎么会.在这里可以学习很多的! 第一天看题自己回答,第二天看牛客网的答案! 1 什么是Java虚拟机?为什么Java被称作是“平台无 ...
- 牛客网《BAT面试算法精品课》学习笔记
目录 牛客网<BAT面试算法精品课>学习笔记 牛客网<BAT面试算法精品课>笔记一:排序 牛客网<BAT面试算法精品课>笔记二:字符串 牛客网<BAT面试算法 ...
- C++版 - HDUoj 2010 3阶的水仙花数 - 牛客网
版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. C++版 - ...
- 牛客网第9场多校E(思维求期望)
链接:https://www.nowcoder.com/acm/contest/147/E 来源:牛客网 题目描述 Niuniu likes to play OSU! We simplify the ...
- 牛客网暑期ACM多校训练营(第七场)Bit Compression
链接:https://www.nowcoder.com/acm/contest/145/C 来源:牛客网 题目描述 A binary string s of length N = 2n is give ...
- Beautiful Numbers(牛客网)
链接:https://ac.nowcoder.com/acm/problem/17385来源:牛客网 题目描述 NIBGNAUK is an odd boy and his taste is stra ...
- 牛客网华为机试题之Python解法
牛客网华为机试题之Python解法 第1题 字符串最后一个单词的长度 a = input().split(" ") print(len(a[-1])) 第2题 计算字符个数 a = ...
- 牛客网Wannafly挑战赛25A 因子(数论 素因子分解)
链接:https://www.nowcoder.com/acm/contest/197/A来源:牛客网 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言5242 ...
- 牛客网 2018年东北农业大学春季校赛 L题 wyh的天鹅
链接:https://www.nowcoder.com/acm/contest/93/L来源:牛客网 时间限制:C/C++ 3秒,其他语言6秒空间限制:C/C++ 262144K,其他语言524288 ...
随机推荐
- Install Ubuntu on Windows Subsystem for Linux
安装参考 ubuntu.com/wsl microsoft/wsl/install-manual microsoft/terminal 错误解决方案 github/启动 WSL 2时警告"参 ...
- 590. N-ary Tree Postorder Traversal - LeetCode
Question 590. N-ary Tree Postorder Traversal Solution 题目大意:后序遍历一个树 思路: 1)递归 2)迭代 Java实现(递归): public ...
- 个人冲刺(三)——体温上报app(二阶段)
冲刺任务:完成用户类.温度数据和第二页面类的编写 User.java package com.example.helloworld; class User { private String usern ...
- Kube-OVN v1.10.0:新增Windows节点支持,用户自定义子网ACL等10+硬核功能
在Kube-OVN社区小伙伴的共同努力下,Kube-OVN v1.10.0于五月份正式发布.Kube-OVN v1.10.0版本中,我们一如既往地对Kube-OVN 的功能.性能.稳定性和易用性进行了 ...
- Pycharm连接远程服务器并保持文件夹同步
pycharm版本2021 服务器版本 Ubuntu 18 1.连接远程服务器 xxx这部省略了,因为之前就已经连接上了hh,后面用到再补充. 2.保持文件夹同步 1.首先在本地(windows环境创 ...
- Oracle数据库丢失表排查思路
Oracle数据库丢失表排查思路 说明:由于系统采用ID取模分表法进行Oracle数据存储,某日发现Oracle数据库中缺少对应的几张业务数据表,遂进行相关问题查询,简单记录一下排查思路: 由于我们代 ...
- .NET C#基础(4):属性 - 本质是方法
0. 文章目的 本文面向有一定.NET C#基础知识的学习者,介绍C#中属性的属性.定义.使用方法以及特殊性. 1. 阅读基础 理解C#基本语法(定义类及类成员,调用方法) 认可OOP的封 ...
- Halodoc使用 Apache Hudi 构建 Lakehouse的关键经验
Halodoc 数据工程已经从传统的数据平台 1.0 发展到使用 LakeHouse 架构的现代数据平台 2.0 的改造.在我们之前的博客中,我们提到了我们如何在 Halodoc 实施 Lakehou ...
- 合宙AIR105(二): 时钟设置和延迟函数
目录 合宙AIR105(一): Keil MDK开发环境, DAP-Link 烧录和调试 合宙AIR105(二): 时钟设置和延迟函数 Air105 的时钟 高频振荡源 芯片支持使用内部振荡源, 或使 ...
- C语言学习之我见-strcmp()字符串比较函数
strcmp()函数,用于两个字符串的比较. (1)函数原型 int strcmp(const char *_Str1,const char *_Str2); (2)头文件 string.h (3)功 ...