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. 【转】c#文件操作大全(一)

    1.创建文件夹//using System.IO;Directory.CreateDirectory(%%1); 2.创建文件//using System.IO;File.Create(%%1); 3 ...

  2. leetcode解题—Longest Palindromic Substring

    题目: Given a string S, find the longest palindromic substring in S. You may assume that the maximum l ...

  3. typedef和#define的区别

    转自:http://www.cnblogs.com/kerwinshaw/archive/2009/02/02/1382428.html 一.typedef的用法在C/C++语言中,typedef常用 ...

  4. WebService 学习过程

    //------------------------------------------------------------------------------------------ //windo ...

  5. GCC交叉编译链命名

    命名格式: arch[-vendor][-os]-abi arch:CPU的架构 vendor:工具链的供应商 os: 目标上运行的操作系统,不同的操作系统对应着不同的C库,例如 newlib.gli ...

  6. C# foreach 原理以及模拟的实现

    public class Person:IEnumerable     //定义一个person类  并且 实现IEnumerable 接口  (或者不用实现此接口 直接在类 //里面写个GetEnu ...

  7. 【已解决】Vmware无法创建虚拟网卡的问题

    最近因为各种需要,要在虚拟机里使用桥接方式连接.但是不管怎么操作,都无法添加虚拟网卡.连续好多天需要用到桥接上网,今儿多方搜索,找到了解决方案. 参考资料:http://tieba.baidu.com ...

  8. javascript高级编程笔记03(正则表达式)

    引用类型 检测数组 注:我们实际开发中经常遇到要把数组转化成以逗号隔开,我以前都是join来实现,其实又更简单的方法可以用toString方法,它会自动用逗号隔开转换成字符串,其实toString内部 ...

  9. VS2015下的Android开发系列01——开发环境配置及注意事项

    概述 VS自2015把Xamarin集成进去后搞Android开发就爽了,不过这安装VS2015完成的时候却是长了不知道多少.废话少说进正题,VS2015安装时注意把Android相关的组件勾选安装, ...

  10. import information website

    1. 一个很好的展示如何review paper和response to reviewer的网站:openview 该网站给出了许多paper review的例子(如何你是reviewer,那么可以参 ...