CF# 260 A. Laptops
2016-02-17 21:32
344 查看
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 test(s)
input
output
cf第一题必定非常水这题,仅仅要读入数据后按价格排序然后找到一对逆序的输出就好了
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 test(s)
input
2 1 2 2 1
output
Happy Alex
cf第一题必定非常水这题,仅仅要读入数据后按价格排序然后找到一对逆序的输出就好了
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> using namespace std; const int maxn=1e5+10; struct node{ int x,y; }e[maxn]; int cmp(node l1,node l2) { return l1.x<l2.x; } int main() { int n; while(~scanf("%d",&n)) { for(int i=0;i<n;i++) scanf("%d%d",&e[i].x,&e[i].y); sort(e,e+n,cmp); int flag=0; for(int i=1;i<n;i++) { if(e[i].x>e[i-1].x&&e[i].y<e[i-1].y) { flag=1; break; } } if(flag) cout<<"Happy Alex"<<endl; else cout<<"Poor Alex"<<endl; } return 0; }
相关文章推荐
- C语言之linux内核--BCD码转二进制与二进制转BCD码(笔试经典)
- C语言之linux内核--BCD码转二进制与二进制转BCD码(笔试经典)
- C语言之linux内核--BCD码转二进制与二进制转BCD码
- opencv 访问图像像素的三种方式
- iTOP-4412 开发板镜像的烧写
- Linux Wget 命令
- shell脚本的#!
- 利用xShell 与linux进行文件互传
- 安卓第二夜 有趣的架构
- Hadoop优化与调整
- nodejs 写bash
- handoop2.0上深度学习解决方案
- C语言之linux内核实现位数高低位互换
- C语言之linux内核实现位数高低位互换
- C语言之linux内核实现位数高低位互换
- linux系统批量格式化磁盘
- CentOS虚拟机环境下安装JDK环境
- ffmpeg文档09-OpenCL选项
- 大型网站架构系列:20本技术书籍推荐
- centos 使用rz sz指令