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.

## Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤$105$). Then the next line contains N distinct positive integers no larger than $109$. The numbers in a line are separated by spaces.

## Output Specification:

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.

5
1 3 2 4 5

3
1 4 5

## 解答

1. 如果一个数能作为切分元素，那它在排序后位置不变，如 1 3 2 4 5，排序后为 1 2 3 4 5， 1 4 5 位置没有变。
2. 假如一个元素的位置是排序后的位置，还不能保证它是切分元素，如 5 2 1 3 4，2显然不是切分元素，因此还要用一个变量max，记录左边元素的最大值，如果当前元素大于左边元素的最大值，那必定大于左边所有元素，且小于右边所有元素。
根据以上两点，算法代码如下（要注意没有切分元素的情况）。
#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;

int main() {
int n;
cin >> n;
vector<int> a(n), b(n);
vector<int> res;
int cnt = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
b[i] = a[i];
}
sort(a.begin(), a.end());
int max = 0;
for (int i = 0; i < n; i++) {
if (a[i] == b[i] && b[i] > max) {
res.push_back(b[i]);
cnt++;
}
if (b[i] > max) max = b[i];
}
cout << cnt << endl;
for (int i = 0; i < cnt - 1; i++) {
cout << res[i] << " ";
}
// 可能会没有主元
if (cnt > 0)
cout << res[cnt - 1];
cout << endl;
return 0;
}