题面

Description

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 $n$ ($1 \le n \le 100$) — 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.

Sample Input 1

3
0 0 1

Sample Output 1

HARD

Sample Input 2

1
0

Sample Output 2

EASY

题解

题目大意

给你一个大小为n的数组,元素只有0和1,如果数组中存在元素为1,则输出HARD,否则输出EASY,并且每个字母大小写随便

分析

直接根据题意模拟,真·签到题

代码

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n = 0;
    while(cin >> n) {
        bool isone = false;
        for(int i = 1; i <= n; i++) {
            int a = 0;

            cin >> a;
            if(a == 1) {
                isone = true;
            }
        }

        if(isone) cout << "HARD" << endl;
        else cout << "EASY" << endl;
    }

    return 0;
}

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