CF 456A(Laptops-排序)

A. Laptops

time limit per test

memory limit per test

input

output

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.

n

Input

n (1 ≤ n ≤ 105) — the number of laptops.

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).

ai are distinct. All bi

Output

Happy Alex", otherwise print "Poor Alex" (without the quotes).

Sample test(s)

input

2 1 2 2 1

output

Happy Alex

先按b排序，再比较a

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
struct node
{
  int a,b;
  friend bool operator<(node a,node b){return a.b<b.b;  }
}a[MAXN];
int n;
int main()
{
  freopen("a.in","r",stdin);
//  freopen("a.out","w",stdout);
  cin>>n;
  For(i,n) scanf("%d%d",&a[i].a,&a[i].b);
  sort(a+1,a+1+n);
  For(i,n-1) if (a[i].a>a[i+1].a) {cout<<"Happy Alex"<<endl; return 0;  }
  
  
  cout<<"Poor Alex"<<endl;
  return 0;
}