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

1
2
3
4
5
6
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

1
2
3
4
5
YES
NO
NO
YES
NO

代码实现

C语言

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#include<stdio.h>

int main() {
	// Max堆栈已压入的最大值，Push、Pop标记压入和取出次数 
	int M, N, K, Max, Push, Pop; 

	scanf("%d %d %d", &M, &N, &K);
	int A[N];	// C99
	for ( int i=0; i<K; i++ ) {		// 有K个序列需要检查 
		Max = Push = Pop = 0;	
		for ( int j=0; j<N; j++ ) {	// 保存序列到数组 
			scanf("%d", &A[j]);
		}
		for ( int m=0; m<N; m++ ) {	// 遍历每个数组 
			if ( A[m] > Max ) {	//  如果读取的数大于记录的最大值，则堆栈中有数字压入 
				Push += A[m] - Max;	// 更新压入的次数 
				if ( Push - Pop <= M ) {	// 压入后堆栈中的数不能大于堆栈能容纳的最大值 
					Pop++;	// 更新取出次数 
					Max = A[m];	// 更新已压入最大数 
				} else {
					break;
				}				
			} else if ( A[m] < Max ) {	// 如果读取的数小于最大值，则堆栈处于取出状态 
				if ( A[m] < A[m-1] ) {	// 处于取出状态时，因为从小到大压入，所以后一个取出的数字一定小于前一个 
					Pop++;
				} else {
					break;
				}
			}
		}
		if ( Push == Pop && Pop == N ) {	// 如果记录的push和pop次数相等，并且等于数字个数，则该序列可输出 
			printf("YES\n");
		} else {
			printf("NO\n");
		}
	}
	
	return 0;
}