PAT A1101.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.

  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:

  5
  1 3 2 4 5
  

  Sample Output:

  3
  1 4 5

  /*********************侯尼玛*********************/

  也是很简单的一题，总体的思路是遍历两次，确定每个点左侧的最大值和右侧的最小值然后进行比较。

  非常奇怪的是一直有一个测试案例通不过，显示格式错误，莫名其妙。大概因为是新题，所以牛客网里找不着，我只能对比网络上搜到的答案，发现在末尾还得加上一个回车才能通过那个测试例的格式……但是其他的测试例怎么就通过了呢？？

  附上代码：

  #include <iostream>
  using namespace std;
  const int LARGE = 0x3FFFFFFF;
  const int SMALL = -0x3FFFFFFF;
  
  int main()
  {
      int N=0;
      cin>>N;
      int *LMax = new int [N+1];
      int *RMin = new int [N+1];
      int *List = new int [N+1];
      for(int i=1;i<=N;i++)
          cin>>List[i];
      RMin[N]=LARGE;
      LMax[1]=SMALL;
      for(int i=2;i<=N;i++)
      {
          if(List[i-1]>LMax[i-1])
              LMax[i]=List[i-1];
          else LMax[i]=LMax[i-1];
      }
      for(int i=N-1;i>=0;i--)
      {
          if(List[i+1]<RMin[i+1])
              RMin[i]=List[i+1];
          else RMin[i]=RMin[i+1];
      }
      int num =0;
      int *PrintList = new int [N+1];
      for(int i=1;i<=N;i++)
      {
          if(LMax[i]<List[i] && RMin[i]>List[i])
          {
              PrintList[num]=List[i];
              num++;
          }
      }
      cout<<num<<endl;;
      for(int i=0;i<num;i++)
      {
          cout<<PrintList[i];
          if(i<num-1) cout<<' ';
      }
      cout<<endl;//不知道为什么，如果没有写这一行的话，始终会有一个测试例显示格式错误。
      return 0;
  }