PTA甲级 1051 Pop Sequence (25分)

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title: PTA 1051 Pop Sequence (25分) date: 2020-10-02 15:43:59 tags:

栈 categories:[刷题,PAT]

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文章目录

题目原文Input Specification:Output Specification:Sample Input:Sample Output: 题目大意:柳神的思路如下:柳神的代码如下: 强烈推荐,刷PTA的朋友都认识一下柳神–PTA解法大佬

本文由参考于柳神博客写成

柳神的博客,这个可以搜索文章

柳神的个人博客,这个没有广告,但是不能搜索

PS 今天也要加油鸭

题目原文

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

题目大意:

就是给你一个长度为M的栈

然后给你一个固定的1要N的入栈顺序,问你能不能得出特定的出栈顺序.

柳神的思路如下:

分析：按照要求进行模拟。先把输入的序列接收进数组v。然后按顺序1~n把数字进栈，每进入一个数字，判断有没有超过最大范围，超过了就break。如果没超过，设current = 1，从数组的第一个数字开始，看看是否与栈顶元素相等，while相等就一直弹出栈，不相等就继续按顺序把数字压入栈～～最后根据变量flag的bool值输出yes或者no～

柳神的代码如下:

#include <iostream> #include <stack> #include <vector> using namespace std; int main() { int m, n, k; scanf("%d %d %d", &m, &n, &k); for(int i = 0; i < k; i++) { bool flag = false; stack<int> s; vector<int> v(n + 1); for(int j = 1; j <= n; j++) scanf("%d", &v[j]); int current = 1; for(int j = 1; j <= n; j++) { s.push(j); if(s.size() > m) break; while(!s.empty() && s.top() == v[current]) { s.pop(); current++; } } if(current == n + 1) flag = true; if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }

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