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The original question is as follows

1001. Rails Time Limit: 1sec Memory Limit:32MB

Description

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track. The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, …, N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, …, aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

Input

The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, …, N. The last line of the block contains just 0. The last block consists of just one line containing 0.

Output

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null’’ block of the input.

Sample Input Copy

5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0

Copy

Sample Output Copy

Yes
No

Yes

Copy

Ideas

Simply put, the carriage comes from the right , Get into Station, And you have to queue up , in other words , If there are several cars in the station , Then the carriage that enters later leaves the station first . This last in first out is just in line with the characteristics of the stack , Consider using the stack to solve . To determine whether a sequence is possible , We just need to see if we can find an inbound and outbound solution to meet this sequence , That is, you only need to test whether the input sequence can go through the operation of putting on the stack and putting out the stack , Get the result sequence . Put the result sequence to be judged into a vector in （ Array equivalent ）, Put the number of stops in the station into stack in . (1) When the station is empty , Must be put on the stack , At this point, if the top of the stack is the same as the first number of the sequence , Out of the stack , And will vector Delete an element , Indicates that a match has been made ; (2) If not , Then continue to stack , Until it matches , When the stack comes out, you must also consider whether the top of the stack can follow vector Match an element of . (3) repeat (1)(2) The operation of . If in the end vector The element of is empty , That is, all elements can be matched , It shows that this sequence is possible . If all the car numbers have been put in the stack , And the top of the stack cannot be connected with vector matching , And vector Not matched yet , It means that this sequence is impossible , Because we can't find a solution to get this sequence .

Code

#include <iostream>
#include <stack>
#include <vector>

using namespace std;

int main(){
	int n;
	cin >> n;
	while(n > 0){
		int a;
		cin >> a;
		if(a == 0) {
			cout << endl;
			cin >> n;
			continue;
		}
		vector<int> num;
		num.push_back(a);
		for(int i = 1; i < n; i++){
			int input;
			cin >> input;
			num.push_back(input);
		}
		int cur = 1;
		stack<int> s;
		/******* The following is the key code ********/
		while(!num.empty()){
			if(cur > n){
				cout << "No" << endl;
				break;
			} 
			s.push(cur);
			cur += 1;
			while(!s.empty() && s.top() == num.front()){
				s.pop();
				num.erase(num.begin());
			}
		}
		if(num.empty()) cout << "Yes" << endl;
	}
}

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