A. Devu, the Singer and Churu, the Joker

最新推荐文章于 2020-10-24 20:57:02 发布

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最新推荐文章于 2020-10-24 20:57:02 发布

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Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.

Devu has provided organizers a list of the songs and required time for singing them. He will sing n songs, ith song will take ti minutes exactly.

The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.

People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.

You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:

The duration of the event must be no more than d minutes;
Devu must complete all his songs;
With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.

If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.

Input

The first line contains two space separated integers n, d (1 ≤ n ≤ 100; 1 ≤ d ≤ 10000). The second line contains n space-separated integers: t1, t2, ..., tn (1 ≤ ti ≤ 100).

Output

If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.

   Sample test(s) 
 
     input 
   
3 30
2 2 1

     output 
   
5

     input 
   
3 20
2 1 1

     output 
   
-1

Note

Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:

First Churu cracks a joke in 5 minutes.
Then Devu performs the first song for 2 minutes.
Then Churu cracks 2 jokes in 10 minutes.
Now Devu performs second song for 2 minutes.
Then Churu cracks 2 jokes in 10 minutes.
Now finally Devu will perform his last song in 1 minutes.

Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.

Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.

解题说明：此题是简单的模拟题，只需要控制好总时间不超过阈值即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include <algorithm>
#include<cstring>
#include<string>
using namespace std;

int main()
{
	int n, d,a[101];
	int i,total,left,sum;
	scanf("%d %d", &n, &d);
	total = 0;
	for (i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
		total += a[i];
		total += 10;
	}
	total -= 10;
	if (total>d)
	{
		printf("-1\n");
	}
	else
	{
		left = d - total;
		sum = left / 5;
		sum += 2 * (n - 1);
		printf("%d\n", sum);
	}
	return 0;
}