PAT-A-1051. Pop Sequence (25)
2017-05-13 23:13
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1051. Pop Sequence (25)
时间限制100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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<iostream> #include<cstdio> #include<stack> using namespace s 4000 td; const int maxn = 1010; int arr[maxn]; stack<int> st; int main() { int m, n, t; cin >> m >> n >> t; while (t--) { while (!st.empty()) st.pop(); for (int i = 1; i <= n; i++) cin >> arr[i]; int current = 1; int flag = 1; for (int i = 1; i <= n; i++) { st.push(i); if (st.size() > m) { flag = 0; break; } while (!st.empty() && st.top() == arr[current]) { st.pop(); current++; } } if (st.empty() == true && flag == 1) cout << "YES" << endl; else cout << "NO" << endl; } system("pause"); return 0; }
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