• Codeforces 456A


    题目链接:http://codeforces.com/problemset/problem/456/A

    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).

    Examples
    Input
    2
    1 2
    2 1
    Output
    Happy Alex

    题解:水水
     1 #include <iostream>
     2 #include <string>
     3 #include <cstring>
     4 #include <algorithm>
     5 #include <cmath>
     6 #include <map>
     7 #include <set>
     8 using namespace std;
     9 #define N 100010
    10 struct com
    11 {
    12     int x,y;
    13 }a[N];
    14 bool cmp(com l,com m){
    15     if(l.x==m.x) return l.y>m.y;
    16     else return l.x<m.x;
    17 }
    18 int main()
    19 {
    20     int n;
    21     while(cin>>n){
    22         for(int i=0;i<n;i++){
    23             cin>>a[i].x>>a[i].y;
    24         }
    25         sort(a,a+n,cmp);
    26         int flag=0;
    27         for(int i=1;i<n;i++){
    28             if(a[i].x>a[i-1].x&&a[i].y<a[i-1].y){
    29                 flag=1;
    30             }
    31             if(flag) break;
    32         }
    33         if(flag) cout<<"Happy Alex"<<endl;
    34         else cout<<"Poor Alex"<<endl;
    35     }
    36     return 0;
    37 }
  • 相关阅读:
    程序员应该看的书籍列表
    完整版QQ(腾讯)开放平台操作指南(包含:qq登录能力获取等等)
    使用Electron构建跨平台的抓取桌面程序
    LinqHelper拓展
    20161014001 DataGridView 单元格内容 自动计算
    20161013001 DataGridView 数据转 DataTable
    20161011001 treeView 递归
    20160929001 Guid生成
    20160815001
    20160715001
  • 原文地址:https://www.cnblogs.com/wydxry/p/7475490.html
Copyright © 2020-2023  润新知