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社区首页 >专栏 >【Codeforces Round】351A - Bear and Game(水)

【Codeforces Round】351A - Bear and Game(水)

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FishWang
发布2025-08-27 08:51:59
发布2025-08-27 08:51:59
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题目链接:点击打开题目

A. Bear and Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bear Limak likes watching sports on TV. He is going to watch a game today.The game lasts 90 minutes and there are no breaks.

Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.

You know that there will be n interesting minutes t1, t2, ..., tn. Your task is to calculate for how many minutes Limak will watch the game.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 90) — the number of interesting minutes.

The second line contains n integers t1, t2, ..., tn (1 ≤ t1 < t2 < ... tn ≤ 90), given in the increasing order.

Output

Print the number of minutes Limak will watch the game.

Examples

input

代码语言:javascript
复制
3
7 20 88

output

代码语言:javascript
复制
35

input

代码语言:javascript
复制
9
16 20 30 40 50 60 70 80 90

output

代码语言:javascript
复制
15

input

代码语言:javascript
复制
9
15 20 30 40 50 60 70 80 90

output

代码语言:javascript
复制
90

Note

In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.

In the second sample, the first 15 minutes are boring.

In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game.

题目不难,上来做题肚子疼也没看清题就开始了,没注意红字的部分。给了俩WA。

代码如下:

代码语言:javascript
复制
#include <cstdio>
int main()
{
	int n;
	int a[100];
	int ans;
	while (~scanf ("%d",&n))
	{
		for (int i = 1 ; i <= n ; i++)
			scanf ("%d",&a[i]);
		ans = 15;
		int i;
		for (i = 1 ; i <= n ; i++)
		{
			if (a[i] <= ans)
				ans = a[i] + 15;
			else
				break;
		}
		if (ans > 90)
			ans = 90;
		printf ("%d\n",ans);
	}
	return 0;
}
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原始发表:2016-05-08,如有侵权请联系 cloudcommunity@tencent.com 删除

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