Codeforces Round #512(Technocup 2019 Elimination Round 1) A. In Search of an Easy (Codeforces 1030A)

版权声明：欢迎转载欢迎评论！ https://blog.csdn.net/rabbit_ZAR/article/details/82827442

题目：In Search of an Easy

题意：给出一个01串，如果全0则输出EASY，否则输出HARD。

思路：入门题……循环下就好

代码：

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

#define ll long long
#define inf (1<<30)
#define read(x) scanf("%d",&x)
#define maxn 200000

int n;
int a[maxn+5];

bool readin() {
	read(n);
	for(int i=1; i<=n; i++) {
		read(a[i]);
		if(a[i]) return true;
	}
	return false;
}

int main() {
	int ans=readin();
	if(ans) printf("HARD");
	else printf("EASY");
	return 0;
}