zoukankan      html  css  js  c++  java
  • 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:
    json_encode不编码中文字符的方式
    网站备份脚本
    英语动词大全
    多线程和多进程的区别【转载网络】
    解决curl中errno为51和60的错误
    ps修改图片文字
    如何设计充值消费的数据表
    LoRa与NB-IoT对比(转载)
    vuejs 使用vue-cli引入bootstrap
    关于防火墙的规则
  • 原文地址:https://www.cnblogs.com/Penn000/p/5758168.html
Copyright © 2011-2022 走看看