1101. Quick Sort (25)

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.

Sample Input:

51 3 2 4 5

Sample Output:

31 4 5

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找出一个序列中所有的轴值，轴值就是比左边的数都要大，比右边的数都要小。所以构建两个数组Max和Min，对于某个位置，求出该位置和其左边的值的最大值，和该位置和其右边的值的最小值，如果该位置的值等于该位置的最大值和最小值，说明就是轴值，将轴值保存在一个数组中。要注意如果轴值数组为空时，第二行不能为空，要输出空行，所以无论什么情况最后都输出空行即可。

代码：

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>#include <algorithm>using namespace std;int main(){int n;cin>>n;vector<int>a(n);vector<int>Min(n);vector<int>Max(n);for(int i=0;i<n;i++){cin>>a[i];if(i==0) {Max[i]=a[i];}else{Max[i]=max(a[i],Max[i-1]);}}for(int i=n-1;i>=0;i--){if(i==n-1){Min[i]=a[i];}else{Min[i]=min(a[i],Min[i+1]);}}vector<int>res;for(int i=0;i<n;i++){if(a[i]==Min[i]&&a[i]==Max[i]){res.push_back(a[i]);}}cout<<res.size()<<endl;for(int i=0;i<res.size();i++){if(i==0) cout<<res[i];else cout<<" "<<res[i];}cout<<endl;}