CF# 260 A. Laptops

全栈程序员-用户IM • 2022年1月27日下午2:00 • 未分类

大家好，又见面了，我是全栈君。

One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.

Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.

Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).

All ai are distinct. All bi are distinct.

Output

If Alex is correct, print “Happy Alex“, otherwise print “Poor Alex” (without the quotes).

         Sample test(s)
       

           input
         
2
1 2
2 1

           output
         
Happy Alex

          cf第一题必定非常水这题，仅仅要读入数据后按价格排序然后找到一对逆序的输出就好了
         
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=1e5+10;
struct node{
    int x,y;
}e[maxn];
int cmp(node l1,node l2)
{
    return l1.x<l2.x;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&e[i].x,&e[i].y);
        sort(e,e+n,cmp);
        int flag=0;
        for(int i=1;i<n;i++)
        {
            if(e[i].x>e[i-1].x&&e[i].y<e[i-1].y)
            {
                flag=1;
                break;
            }
        }
        if(flag)
            cout<<"Happy Alex"<<endl;
        else
            cout<<"Poor Alex"<<endl;
    }
    return 0;
}