这是本人第一次写代码,难免有点瑕疵还请见谅

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 nd (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.

关于这题的解题报告:本人觉得当你想多了的时候就会把太多的时间浪费掉了,因为本人就体验过,算了发了一下牢骚就打住吧;

对于这题:基本思路就是简单的模拟;

下面就是本人的代码:

 #include<cstdio>
#define N 100 int t[N];
int main()
{
int d,n;
int Sum = ;
int count = ; scanf("%d%d",&n,&d);
for(int i=;i<=n;++i){
scanf("%d",&t[i]);
Sum += t[i];
}
Sum += (n-)*;
if(Sum > d)
printf("-1\n");
else
{
count += (n-)*;
printf("%d\n",count+(d-Sum)/);
}
int a;scanf("%d\n",&a);
return ;
}

Codeforces 439 A. Devu, the Singer and Churu, the Joker的更多相关文章

  1. Codeforces Round #251 (Div. 2) A - Devu, the Singer and Churu, the Joker

    水题 #include <iostream> #include <vector> #include <algorithm> using namespace std; ...

  2. codeforces 439 E. Devu and Birthday Celebration 组合数学 容斥定理

    题意: q个询问,每一个询问给出2个数sum,n 1 <= q <= 10^5, 1 <= n <= sum <= 10^5 对于每一个询问,求满足下列条件的数组的方案数 ...

  3. codeforces#439 D. Devu and his Brother (二分)

    题意:给出a数组和b数组,他们的长度最大1e5,元素范围是1到1e9,问你让a数组最小的数比b数组最大的数要大需要的最少改变次数是多少.每次改变可以让一个数加一或减一 分析:枚举a数组和b数组的所有的 ...

  4. Codeforces 493 E.Devu and Birthday Celebration

    \(>Codeforces \space 493\ E.Devu\ and\ Birthday\ Celebration<\) 题目大意 : 有 \(q\) 组询问,每次有 \(n\) 小 ...

  5. Codeforces #439 Div2 E

    #439 Div2 E 题意 给出二维平面,有多个询问: 把某一区域围起来(围墙之间无交点) 移除某一区域的围墙(此时保证围墙一定存在) 选定两个位置问是否可以互相到达 分析 看起来很复杂,其实这道题 ...

  6. Codeforces 451 E Devu and Flowers

    Discription Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th ...

  7. 【Codeforces 258E】 Devu and Flowers

    [题目链接] http://codeforces.com/contest/451/problem/E [算法] 容斥原理 [代码] #include<bits/stdc++.h> usin ...

  8. Codeforces 451 E. Devu and Flowers(组合数学,数论,容斥原理)

    传送门 解题思路: 假如只有 s 束花束并且不考虑 f ,那么根据隔板法的可重复的情况时,这里的答案就是 假如说只有一个 f 受到限制,其不合法时一定是取了超过 f 的花束 那么根据组合数,我们仍然可 ...

  9. Codeforces Round 251 (Div. 2)

    layout: post title: Codeforces Round 251 (Div. 2) author: "luowentaoaa" catalog: true tags ...

随机推荐

  1. [HTML]marquee标签属性详解

    请大家先看下面这段代码:<marquee direction=upbehavior=scrollloop=3scrollamount=1 scrolldelay=10align=topbgcol ...

  2. C# double float int string 与 byte数组 相互转化

    在做通信编程的时候,数据发送多采用串行发送方法,实际处理的时候多是以字节为单位进行处理的.在C/C++中 多字节变量与Byte进行转化时候比较方便 采用UNION即可废话少说看示例:typedef u ...

  3. Redhat 6 配置CentOS yum source

    由于最近曝出linux的bash漏洞,想更新下bash,于是 想到了配置CentOS yum source. 测试bash漏洞的命令: env x='() { :;}; echo "Your ...

  4. CDH安装Hadoop

    一.安装CDH-manager 1.关闭selinux 修改/etc/selinux/config 文件 将SELINUX=enforcing改为SELINUX=disabled 重启机器即可   2 ...

  5. 如何做到 jQuery-free?

    一.选取DOM元素 jQuery的核心是通过各种选择器,选中DOM元素,可以用querySelectorAll方法模拟这个功能. var $ = document.querySelectorAll.b ...

  6. TCP UDP 协议的选择

    行业应用中TCP/IP传输协议和UDP协议的选择! 中国移动.中国联通推行的GPRS网络.CDMA网络已覆盖大量的区域,通过无线网络实现数据传输成为可 能.无线Modem采用GPRS.CDMA模块通过 ...

  7. org.springframework.orm.jpa.JpaTransactionManager

    [第九章] Spring的事务 之 9.2 事务管理器 ——跟我学spring3 http://sishuok@com/forum/blogPost/list/0/2503.html

  8. Elasticsearch基础概念理解

    熟悉ES中的几个关键概念: 节点(Node):一个elasticsearch运行的实例,其实就是一个java进程.一般情况下,一台机器运行在一台机器上. 集群(Cluster): 好几个有相同集群名称 ...

  9. Android手势监听

    public class MainActivity extends Activity { /* * 要实现手指在屏幕上左右滑动的事件需要实例化对象GestureDetector,new * Gestu ...

  10. php nl2br() 函数

    nl2br() 函数在字符串中的每个新行 (\n) 之前插入 HTML 换行符 (<br />).