Click on the link full solution summary PAT B -AC
Title:
The famous fast sorting algorithm in a classic division of the process: We usually use some method takes an element as a principal component, through the exchange, the yuan less than the main elements of it into the left than the main elements of Yuan put it right. The arrangement of the N positive integers different from each other after a given division, PCA may I ask how many elements can be divided into pre-selected?
For example given $ N = 5 $, is arranged 1,3,2,4,5. then:
1 left no elements, the elements on the right than it is large, so it could be PCA;
although left element 3 than it is small, but its right than two hours it, so it can not be a principal component;
although 2 the right elements than it is big, but big 3 on its left than it is, so it can not be a principal component;
for similar reasons, 4 and 5 are likely to be the main yuan.
Accordingly, three elements may be Principal Component Analysis.
Input format:
Enter a given positive integer N (1 ≦ 10 in line 5 ); the second row is separated by a space N different positive integers, the number of not more than 10 . 9 .
Output format:
in the first line is the number of possible output element of the main elements; output in ascending order of these elements in the second row in the system, separated by a space, the line may not have extra space inclusive.
Sample input:
5
1 3 2 4 5
Sample output:
3
1 4 5
My code:
#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstring>
#include<sstream>
using namespace std;
//有的时候题目是一起做的，所以会有不需要的头文件
int main()
{
int N;
scanf("%d\n",&N);
set<int>res;
int biggest=0;
for(int i=0;i<N;i++)
{
int temp;
cin>>temp;
if(temp>biggest)
{
res.insert(temp);
biggest=temp;
}
else
{
for(set<int>::iterator it=res.begin();it!=res.end();it++)
{
if(*it>temp)
{
res.erase(it,res.end());
break;
}
}
}
}
if(res.size()==0)
{
cout<<"0"<<endl<<endl;
return 0;
}
cout<<res.size()<<endl;
for(set<int>::iterator it=res.begin();it!=res.end();it++)
{
if(it==res.begin())cout<<*it;
else cout<<" "<<*it;
}
return 0;
}
case2 been reported format error ... the original ...
