zoukankan      html  css  js  c++  java
  • CF# 260 A. Laptops

    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
    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;
    }


  • 相关阅读:
    window.location.reload();页面实现跳转和刷新
    vue自定义指令--directive
    1019 数字黑洞
    1018 锤子剪刀布
    1017 A除以B
    1016 部分A+B
    1015 德才论
    1014 福尔摩斯的约会
    1013 数素数
    1012 数字分类
  • 原文地址:https://www.cnblogs.com/yxwkf/p/5196582.html
Copyright © 2011-2022 走看看