*Time limit ：1000 ms
Memory limit ：262144 kB
Source ：Codeforces Round #251 (Div. 2)
Tags ：greedy implementation 1100
Editorial ：Announcement Tutorial
问题描述 ：
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.
Examples
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.
#include <iostream>
using namespace std;
int main()
{
int n, d, i = 0, k = 0,s=0;
cin >> n >> d;
int *t = new int[n];
k = 10 * (n - 1);
s = k / 5;
for (i = 0; i < n; i++)
{
cin >> t[i];
k += t[i];
}
if (k > d)
cout << -1 << endl;
else
{
if (d - k >= 5)
cout << s+(d-k)/5<< endl;
else
cout <<s << endl;
}
delete []t;
return 0;
}