【cf439A】Devu, the Singer and Churu, the Joker

描述 Description

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.

输入格式 InputFormat

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).

输出格式 OutputFormat

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.

样例输入 SampleInput

3 30
2 2 1

样例输出 SampleOutput

5

数据范围和注释 Hint

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.

Codeforces 439A

代码 Code

水。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int a[105];
int i,j,t,d,n,m,l,r,k,z,y,x,sum,ans;
int main()
{
    scanf("%d%d",&n,&d);
    sum=0;
    for (i=1;i<=n;i++) 
    {
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    if (sum+(n-1)*10>d)
    {
        printf("-1\n");
        return 0;
    }
    ans=0;
    ans=(d-sum)/5;
    printf("%d\n",ans);
    return 0;
}