# 02-3. 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 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 = 1000;

int main()
{
int m, n, k;
int arr[maxn];
scanf("%d%d%d", &m, &n, &k);
for(int i = 0; i < k; i++)
{
int A = 1, B = 0;
int ok = 1;
stack<int> s;
for(int j = 0; j < n; j++)
{
scanf("%d", &arr[j]);
}
while(B < n)
{
if(A == arr[B] && s.size() < m)
{
A++;
B++;
}
else if(!s.empty() && s.top() == arr[B])
{
s.pop();
B++;
}
else if(A <= n && s.size() < m)
{
s.push(A);
A++;
}
else
{
ok = 0;
break;
}
}
printf("%s\n", ok ? "YES" : "NO");
}
return 0;
}</span>

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