Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO
#include<cstdio>
#include<stack>
using namespace std;
const int maxn = ;
int a[maxn];
stack<int> s;
int main(){
int m,n,k;
scanf("%d%d%d",&m,&n,&k);
while(k--){
while(!s.empty()){
s.pop();
}
for(int i = ; i <= n; i++){
scanf("%d",&a[i]);
}
int current = ;
bool flag = true;
for(int i = ; i <= n; i++){
s.push(i);
if(s.size() > m){
flag = false;
break;
}
while(!s.empty() && a[current] == s.top()){
s.pop();
current++;
}
}
if(s.empty() && flag == true) printf("YES\n");
else printf("NO\n");
}
return ;
}

02-线性结构4 Pop Sequence (25 分)的更多相关文章

  1. PTA 02-线性结构4 Pop Sequence (25分)

    题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/665 5-3 Pop Sequence   (25分) Given a stack wh ...

  2. 02-线性结构4 Pop Sequence (25 分)

    Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and p ...

  3. pat02-线性结构4. Pop Sequence (25)

    02-线性结构4. Pop Sequence (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue Given ...

  4. PAT 甲级 1051 Pop Sequence (25 分)(模拟栈,较简单)

    1051 Pop Sequence (25 分)   Given a stack which can keep M numbers at most. Push N numbers in the ord ...

  5. PAT 1051 Pop Sequence (25 分)

    返回 1051 Pop Sequence (25 分)   Given a stack which can keep M numbers at most. Push N numbers in the ...

  6. 线性结构4 Pop Sequence

    02-线性结构4 Pop Sequence(25 分) Given a stack which can keep M numbers at most. Push N numbers in the or ...

  7. 数据结构练习 02-线性结构3. Pop Sequence (25)

    Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and p ...

  8. 浙大数据结构课后习题 练习二 7-3 Pop Sequence (25 分)

    Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and p ...

  9. 1051 Pop Sequence (25分)

    Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and p ...

  10. 【PAT甲级】1051 Pop Sequence (25 分)(栈的模拟)

    题意: 输入三个正整数M,N,K(<=1000),分别代表栈的容量,序列长度和输入序列的组数.接着输入K组出栈序列,输出是否可能以该序列的顺序出栈.数字1~N按照顺序随机入栈(入栈时机随机,未知 ...

随机推荐

  1. [Token] 从index.jsp中获取Token

    import com.eviware.soapui.support.GroovyUtils def groovyUtils = new GroovyUtils( context ) def holde ...

  2. 【转载】Redis优化经验

    转载地址:http://blog.sina.com.cn/s/blog_4be888450100z2ze.html 内存管理优化 Redis Hash是value内部为一个HashMap,如果该Map ...

  3. jQuary总结7:动画操作,显示与隐藏 淡入淡出, 滑入滑出

    1 jquery提供了三组基本动画,这些动画都是标准的.有规律的效果,jquery还提供了自定义动画的功能. 2 显示与隐藏: show([speed],[easing],[callback]) 显示 ...

  4. unity 小地图的制作

    利用 Transform.InverseTransformDirection  变换位置从世界坐标到自身坐标. 以第一人称控制器为坐标原点(忽视y轴),x轴z轴转为屏幕坐标. 若物体在地图范围外,可以 ...

  5. Head First Python之3文件与异常

    文件基本操作 Python从文本读取数据时,一次会到达一个数据行. sketch.txt文件 Man: Is this the right room for an argument? Other Ma ...

  6. 【Linux】GCC编译器

    [简介] GCC是Linux下的编译工具集,是GNU Compiler Collection的缩写,包含gcc g++ 等编译器.GCC工具集不仅能编译C/C++语言,其他例如Object-c.Pas ...

  7. LRU缓存介绍与实现 (Java)

    引子: 我们平时总会有一个电话本记录所有朋友的电话,但是,如果有朋友经常联系,那些朋友的电话号码不用翻电话本我们也能记住,但是,如果长时间没有联系 了,要再次联系那位朋友的时候,我们又不得不求助电话本 ...

  8. Autoconf 中文手册

    Autoconf Autoconf Creating Automatic Configuration Scripts Edition 2.13, for Autoconf version 2.13 D ...

  9. Java知多少虚拟机(JVM)以及跨平台原理

    相信大家已经了解到Java具有跨平台的特性,可以“一次编译,到处运行”,在Windows下编写的程序,无需任何修改就可以在Linux下运行,这是C和C++很难做到的. 那么,跨平台是怎样实现的呢?这就 ...

  10. JavaScript的词法作用域问题

    多年以前,当我怀揣着前端工程师的梦想时,曾经认真阅读过<JavaScript高级程序设计(第2版)>.里面有一个问题(P147),让我一直百思不得其解. function createFu ...