In Search of an Easy Problem

文章目录

一、In Search of an Easy Problem

总结

一、In Search of an Easy Problem

本题链接：In Search of an Easy Problem

题目：

A. In Search of an Easy Problem

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

When preparing a tournament, Codeforces coordinators try treir best to make the first problem as easy as possible. This time the coordinator had chosen some problem and asked n people about their opinions. Each person answered whether this problem is easy or hard.

If at least one of these n people has answered that the problem is hard, the coordinator decides to change the problem. For the given responses, check if the problem is easy enough.

Input

The first line contains a single integer n (1≤n≤100) — the number of people who were asked to give their opinions.

The second line contains n integers, each integer is either 0 or 1. If i-th integer is 0, then i-th person thinks that the problem is easy; if it is 1, then i-th person thinks that the problem is hard.

Output

Print one word: “EASY” if the problem is easy according to all responses, or “HARD” if there is at least one person who thinks the problem is hard.

You may print every letter in any register:"EASY", "easy", "EaSY" and "eAsY" all will be processed correctly.

Examples

input

3

0 0 1

output

HARD

input

1

0

output

EASY

Note

In the first example the third person says it’s a hard problem, so it should be replaced.

In the second example the problem easy for the only person, so it doesn’t have to be replaced.

本博客给出本题截图：

题意：输入n个数，0代表简单，1代表难，一旦有一个人认为是难的，那就输出HARD，否则输出EASY.

AC代码

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for (int i = 0; i < n; i ++ )
    {
        int a;
        cin >> a;
        if (a)
        {
            puts("HARD");
            exit(0);
        }
    }
    puts("EASY");
    return 0;
}

总结

水题，不解释