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PAT甲级——A1051 Pop Sequence
阅读量:4540 次
发布时间:2019-06-08

本文共 1692 字,大约阅读时间需要 5 分钟。

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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO
1 #include 
2 #include
3 using namespace std; 4 int M, N, K; 5 int main() 6 { 7 cin >> M >> N >> K; 8 for (int i = 0; i < K; ++i) 9 {10 stack
s;11 int a, b = 1, flag = 1;12 for (int j = 0; j < N; ++j)13 {14 cin >> a;15 while (s.size() < M && (s.empty() || s.top() != a))16 s.push(b++);17 if (s.top() == a)18 s.pop();19 else20 flag = 0;21 }22 if (flag == 1)23 cout << "Yes" << endl;24 else25 cout << "No" << endl;26 }27 return 0;28 }

 

转载于:https://www.cnblogs.com/zzw1024/p/11280809.html

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