CodeForces 673 A Bear and Game

本题描述了一只喜欢观看体育赛事的熊Limak,在面对一场90分钟的比赛时的行为模式。如果连续15分钟的比赛无聊,Limak会关闭电视。题目提供了一系列有趣的时间点,要求计算Limak实际观看比赛的时间。

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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
3
7 20 88
output
35
input
9
16 20 30 40 50 60 70 80 90
output
15
input
9
15 20 30 40 50 60 70 80 90
output
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.

题意: 给出一串数字表示在这90分钟内的精彩时刻,其他时间都为无聊时间。假如无聊时间持续大于15分钟,那么他就会关掉电视。问你他最长能看多长时间的电视。

AC代码:

#include<stdio.h>
int main()
{
	int n;
	int a[100];
	while(scanf("%d",&n)!=EOF)
	{
		int falg=0;
		for(int i=0;i<n;i++)
		    scanf("%d",&a[i]);
		if(a[0]>15)
		{
			printf("15\n");
			continue;
		}	   
		else
	    {
			for(int i=0;i<n-1;i++)
			{	
				if(a[i+1]-a[i]>15)
				{
					printf("%d\n",a[i]+15);
					falg=1;
					break;
				}
			}
			if(falg!=1)
			{
				if(90-a[n-1]>15)
				   printf("%d\n",a[n-1]+15);
				else
				   printf("90\n");
			}
		}
	}
	return 0;
 } 


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