第三次周賽A - Problem A CodeForces - 439A

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.
問題分析:今天才認識到原來在外面定義且賦初值,是可以在下面循環加數且循環外輸出的。這道題沒有考慮到唱歌唱完了但是還有很多時間可以一直講笑話的次數。
問題鏈接:https://vjudge.net/contest/276590#problem/A
代碼:

#include <iostream>
using namespace std;
int main()
{
	int n, d, a[100], sum = 0, k, s;
	cin >> n >> d;
	for (int j = 0; j < n; j++)
	{
		cin >> a[j];
		sum += a[j];
	}
	sum = sum + (n - 1) * 10;
	if (sum > d)
	{
		cout << -1;
	}
	if (sum <= d)
	{
		s = d - sum;
		cout << (n - 1) * 2 + (s / 5);
	}
}
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