A. In Search of an Easy Problem
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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 nn people about their opinions. Each person answered whether this problem is easy or hard.
If at least one of these nn 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 nn (1≤n≤1001≤n≤100) — the number of people who were asked to give their opinions.
The second line contains nn integers, each integer is either 00 or 11. If ii-th integer is 00, then ii-th person thinks that the problem is easy; if it is 11, then ii-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
Copy
3 0 0 1
output
Copy
HARD
input
Copy
1 0
output
Copy
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.
解題說明:此題只需要判斷數列中是否存在1即可。
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
int n, i, a, s = 0;
scanf("%d", &n);
for (i = 1; i <= n; i++)
{
scanf("%d", &a);
s = s + a;
}
if (s <= 0)
{
printf("Easy\n");
}
else
{
printf("Hard\n");
}
return 0;
}