（PAT 1101)Quick Sort (递推法)

There is a classical process namedpartitionin 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. GivenNdistinct positive integers after a run of partition, could you tell how many elements could be the selected pivot for this partition?

For example, givenN=5and 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 integerN(≤105). Then the next line containsNdistinct positive integers no larger than109. 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

解题思路:

这个问题没有必要写一个快速的排列，然后找到一个比较，主题的意思是判断每个元素左边的元素是否小于当前的元素，右边的元素是否大于当前的元素，如果满足，输出元素

这种递推法可以解决，从左到右递推，给出数组目前位置的最大元素

然后从右向左推，给出数组中最小的元素

然后将原数组中的元素依次与递推数组中的元素进行比较。

#include <iostream>#include <algorithm>#include <vector>using namespace std;const int MAXN = 100010;long int numbers[MAXN];long int forwards[MAXN];long int backs[MAXN];int  N;int main() {scanf("%d", &N);for (int i = 0; i < N; ++i) {scanf("%d", &numbers[i]);}//正推int biggest = numbers[0];forwards[0] = numbers[0];for (int j = 1; j < N; ++j) {forwards[j] = forwards[j - 1];  ///推if (numbers[j] > biggest) {biggest = numbers[j];forwards[j] = biggest;}}//逆推int smallest = numbers[N - 1];backs[N-1] = numbers[N - 1];for (int k = N - 2; k >= 0; --k) {backs[k] = backs[k+1];if (numbers[k] < smallest) {smallest = numbers[k];backs[k] = smallest;}}vector<long int> res;for (int i = 0; i < N; ++i) {if (numbers[i] >= forwards[i] && numbers[i] <= backs[i]) {res.push_back(numbers[i]);}}  cout << res.size() << endl;for (int i = 0; i < res.size(); ++i) {cout << res[i];if(i < res.size()-1){  cout<<" ";}}cout<<endl;  //必须换行return 0;}