Problem 1030A In Search of an Easy Problem (Codeforces 800)-优快云博客

本文链接：https://blog.youkuaiyun.com/Technodream/article/details/120627474

协调者在准备比赛时尽力让第一个问题尽可能简单。如果询问的n个人中有至少一个人认为问题困难，就会更换问题。输入n（1≤n≤100）和n个0或1的整数，表示每个人的意见。如果所有人的回答都是0，则输出"EASY"，否则输出"HARD"。

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In Search of an Easy Problem

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 $n$ ( $1 \leq n \leq 100$ ) — 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

Simplified Question 题目大意

输入若干数字, 全部是 $0$ 输出EASY, 有一个 $1$ 就输出HARD.

Accepted Answer AC 代码

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

int main(){
	int n, a; cin>>n;
	while (n--){
		cin>>a;
		if (a==1){
			cout<<"Hard"; return 0; //这里要return 0 结束程序
		}
	}
	cout<<"Easy";
	return 0;
}

AC 证明

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