Codeforces Round #260 (Div. 2)A. Laptops

A. Laptops

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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.

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,
a_i and
b_i
(1 ≤ a_i, b_i ≤ n), where
a_i is the price of the
i-th laptop, and b_i is the number that represents the quality of the
i-th laptop (the larger the number is, the higher is the quality).

All a_i are distinct. All
b_i 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

题目告诉。给出n个电脑，每一个电脑一个价钱和价值，问有没有存在价钱更低。价值更高的，sad。。

。。一開始还想要排序神马的，越想越乱。最后发现 输入的ai和bi都是不同的，而且都在1到n的范围内，也就是说每一个数仅仅会出现一次，那么仅仅有 a[1] =1 a[2] = 2 这样时，才不会出现价钱低而价值高的，哈希一次遍历就能够，sad。。

。。15分钟才过。

。。。

#include <cstdio>
#include <cstring>
#include <cstring>
#define INF 0x3f3f3f3f
using namespace std;
int p[110000] ;
int main()
{
    int i , n , a , b ;
    while(scanf("%d", &n)!=EOF)
    {
        memset(p,-1,sizeof(p));
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%d %d", &a, &b);
            p[a] = b ;
        }
        for(i = 1 ; i <= n ; i++)
            if( p[i] != i )
                break;
        if(i <= n)
            printf("Happy Alex\n");
        else
            printf("Poor Alex\n");

    }
    return 0;
}