1051. Pop Sequence (25)
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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
这个题刚开始完全没有思路,后来看了别人写的博客,发现此题并不难。打个比喻,如果要检查一辆车是否合格,肯定是每个零件都要检查一遍(这个比喻不一定恰当),同样,检查一个序列是否可由条件生成,那么就检查一下这个序列的每个元素是否可生成。所以问题划归为检查一个数。
算法的定义是有限个有序的步骤解决一个问题,所以很多时候,走出第一步就成功了一般。
具体代码如下:
/*************************************************************************> File Name: 1051.c> Author: > Mail: > Created Time: 2016年12月22日 星期四 23时39分48秒 ************************************************************************/#include<stdio.h>#include <malloc.h>int main(){ int M, N, K; int popEle, pushEle; int i, j, k = 1; int rear = -1; int flag = 1; scanf("%d%d%d", &M, &N, &K); int* Stack = (int*)malloc(M * sizeof(int)); for (i = 0; i < K; i++) { flag = 1; rear = -1; k = 1; for (j = 0; j < N; j++) { //inspect j_th element scanf("%d", &popEle); if (rear != -1 && Stack[rear] == popEle) { //find it in the stack rear -= 1; } else { //find in the numbers not pushed into stack while (k != popEle && k <= N && rear < M-2) { //rear < M-2, not M-1 Stack[++rear] = k; ++k; } if (k != popEle) { flag = 0; } else { ++k; } } } if (flag == 1) { printf("YES\n"); } else { printf("NO\n"); } } return 0;}
编程之路,进步,再进一步!
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
- 1051. Pop Sequence (25)
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