1101 Quick Sort

There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. Given N distinct positive integers after a run of partition, could you tell how many elements could be the selected pivot for this partition?

For example, given N=5 and the numbers 1, 3, 2, 4, and 5. We have:

1 could be the pivot since there is no element to its left and all the elements to its right are larger than it;

3 must not be the pivot since although all the elements to its left are smaller, the number 2 to its right is less than it as well;

2 must not be the pivot since although all the elements to its right are larger, the number 3 to its left is larger than it as well;

and for the similar reason, 4 and 5 could also be the pivot.

Hence in total there are 3 pivot candidates.

Each input file contains one test case. For each case, the first line gives a positive integer N (≤). Then the next line contains N distinct positive integers no larger than 1. The numbers in a line are separated by spaces.

For each test case, output in the first line the number of pivot candidates. Then in the next line print these candidates in increasing order. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

題意：

　　給出一組數字，問這組數字中的一個數是否滿足該數字左邊的數都比這個數字小，該數字右邊的數都比這個數字大。

思路：

　　首先構造兩個數組，一個用來存放該數字前面最大的數字，一個用來存放改數字後面最小的數字，然後用原數組與這兩個數組比較，看是否滿足要求。

Code：

永遠渴望，大智若愚（stay hungry, stay foolish）