2018/12/22 acm日常 第三周 CodeForces - 439A

A - Problem A

问题原址

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（注意事项）
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.

源代码

#include<iostream>
using namespace std;
int main()
{
	int n, d, sum, x;
	int t[105] = {0};
	cin >> n >> d;
	for (int i=0;i<n;i++)
	{
		cin >> t[i];
		d = d - t[i];
	}
	sum = d - (n - 1) * 10;
	x = 2 * (n - 1);///玩笑次数
	if (sum<0)
	{
		cout << -1 << endl;
	}
	else
	{
		if (sum >= 5)
		{
			x += sum / 5;
		}
		cout << x << endl;
	}
    return 0;
}

思想：循环输入的同时减去演奏时间，再减去休息时间（必要），看时间是否有剩。
不够则-1。
有多的时间看能讲多少笑话。用 / （除）计算，输出讲笑话次数（休息*2+多余）即可。