Codeforces 439 A. Devu, the Singer and Churu, the Joker，codeforcesdevu

Codeforces 439 A. Devu, the Singer and Churu, the Joker，codeforcesdevu

 這是本人第一次寫代碼，難免有點瑕疵還請見諒                                                                                          

A. Devu, the Singer and Churu, the Joker time limit per test  1 second memory limit per test  256 megabytes input  standard input output  standard output

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.

 

 

關於這題的解題報告：本人覺得當你想多了的時候就會把太多的時間浪費掉了，因為本人就體驗過，算了發了一下牢騷就打住吧；

對於這題：基本思路就是簡單的模擬；

下面就是本人的代碼：

 1 #include<cstdio>
 2 #define N 100
 3 
 4 int t[N];
 5 int main()
 6 {
 7     int d,n;
 8     int Sum = 0;
 9     int count = 0;
10     
11     scanf("%d%d",&n,&d);
12     for(int i=1;i<=n;++i){
13         scanf("%d",&t[i]);
14         Sum += t[i];
15     }
16     Sum += (n-1)*10;
17     if(Sum > d)
18          printf("-1\n");
19     else
20     {
21         count += (n-1)*2;
22         printf("%d\n",count+(d-Sum)/5);
23     }
24     int a;scanf("%d\n",&a);
25     return 0;
26 }