• CodeForces 456A


    Laptops

    Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

    Description

    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 Input

    Input
    2
    1 2
    2 1
    Output
    Happy Alex

     1 //2016.8.2
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 struct laptop
     9 {
    10     int a, b;
    11 }lap[100005];
    12 
    13 bool cmp(laptop x, laptop y)
    14 {
    15     if(x.a == y.a)return x.b < y.b;
    16     return x.a < y.a;
    17 }
    18 
    19 int main()
    20 {
    21     int n;
    22     bool fg;
    23     while(scanf("%d", &n)!=EOF)
    24     {
    25         fg = false;
    26         for(int i = 0; i < n; i++)
    27         {
    28             scanf("%d %d", &lap[i].a, &lap[i].b);
    29         }
    30         sort(lap, lap+n, cmp);
    31 
    32         for(int i = 1; i < n; i++)
    33             if(lap[i].b < lap[i-1].b)
    34             {
    35                 fg = true;
    36                 break;
    37             }
    38         if(fg)cout << "Happy Alex
    ";
    39         else cout << "Poor Alex
    ";
    40     }
    41 
    42     return 0;
    43 }
  • 相关阅读:
    2018 南京网络预赛Sum
    一个莫比乌斯等式的证明
    LOJ 2452 对称 Antisymmetry——用hash求回文串数
    LOJ 103子串查找——用hash代替kmp算法
    LOJ2823 三个朋友 ——查询字串的哈希值
    hash入门
    2019牛客暑期多校训练营(第十场)Coffee Chicken——递归
    2019牛客暑期多校训练营(第十场)Han Xin and His Troops——扩展中国剩余定理
    mutex 的 可重入
    Linux 编译安装Boost
  • 原文地址:https://www.cnblogs.com/Penn000/p/5758168.html
Copyright © 2020-2023  润新知