A. Rewards
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bizon the Champion is called the Champion for a reason.

Bizon the Champion has recently got a present — a new glass cupboard with
n shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has
a1 first prize cups,
a2 second prize cups and
a3 third prize cups. Besides, he has
b1 first prize medals,
b2 second prize medals and
b3 third prize medals.

Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:

  • any shelf cannot contain both cups and medals at the same time;
  • no shelf can contain more than five cups;
  • no shelf can have more than ten medals.

Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.

Input

The first line contains integers a1,
a2 and
a3 (0 ≤ a1, a2, a3 ≤ 100). The second line contains integers
b1,
b2 and b3
(0 ≤ b1, b2, b3 ≤ 100). The third line contains integer
n (1 ≤ n ≤ 100).

The numbers in the lines are separated by single spaces.

Output

Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).

Sample test(s)
Input
1 1 1
1 1 1
4
Output
YES
Input
1 1 3
2 3 4
2
Output
YES
Input
1 0 0
1 0 0
1
Output
NO
水题不解释
#include <iostream>
using namespace std;
int main()
{
int n,a1,a2,a3,b1,b2,b3;
cin>>a1>>a2>>a3>>b1>>b2>>b3>>n;
if(n>=(a1+a2+a3+4)/5+(b1+b2+b3+9)/10)
cout<<"YES";
else
cout<<"NO"; }

版权声明:本文博主原创文章。博客,未经同意不得转载。

codeforces #256 A. Rewards的更多相关文章

  1. Codeforces #256 Div.2

    B. Suffix Structure 1. 先判断s去掉一些元素是否能构成t,如果可以就是automaton 判断的方法也很简单,two pointer,相同元素同时++,不相同s的指针++,如果t ...

  2. codeforces 256 div2 C. Painting Fence 分治

    C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard in ...

  3. Codeforces Round #256 (Div. 2) A. Rewards

    A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input out ...

  4. Codeforces Round #256 (Div. 2) 题解

    Problem A: A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standar ...

  5. Codeforces Round #256 (Div. 2) D. Multiplication Table(二进制搜索)

    转载请注明出处:viewmode=contents" target="_blank">http://blog.csdn.net/u012860063?viewmod ...

  6. Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)

    解题报告 意思就是说有n行柜子,放奖杯和奖牌.要求每行柜子要么全是奖杯要么全是奖牌,并且奖杯每行最多5个,奖牌最多10个. 直接把奖杯奖牌各自累加,分别出5和10,向上取整和N比較 #include ...

  7. CF#256(Div.2) A. Rewards

    A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input out ...

  8. Codeforces Round #256 (Div. 2) B. Suffix Structures(模拟)

    题目链接:http://codeforces.com/contest/448/problem/B --------------------------------------------------- ...

  9. Codeforces Round #256 (Div. 2)

    A - Rewards 水题,把a累加,然后向上取整(double)a/5,把b累加,然后向上取整(double)b/10,然后判断a+b是不是大于n即可 #include <iostream& ...

随机推荐

  1. 查看进程所用的内存(使用GetWindowThreadProcessId取得进程ID,OpenProcess打开进程和GetProcessMemoryInfo取得内存信息)

    // function GetProcessMemorySize(_sProcessName: string; var _nMemSize: Cardinal): Boolean; var l_nWn ...

  2. common lisp wiki

    CLiki: index   http://www.cliki.net/

  3. sql server 2012 数据库还原方法

    USE master RESTORE DATABASE WSS_Content FROM DISK = N'D:\bak\contentbak.bak' WITH REPLACE, NORECOVER ...

  4. 指尖上的电商---(3)Solr全文搜索引擎的配置

    接上篇,Solr的准备工作完毕后,本节主要介绍Solr的安装,事实上Solr不须要安装.直接下载就能够了      1.Solr配置 下载地址 :http://lucene.apache.org/so ...

  5. QLineEdit 自动完成(使用setCompleter,内含一个ListView)

    -------------------------------------CompleteLineEdit.h------------------------------------- #ifndef ...

  6. Threejs 的场景查看 - 几个交互事件库助你方便查看场景

    Threejs 的场景查看 - 几个交互事件库助你方便查看场景 太阳火神的漂亮人生 (http://blog.csdn.net/opengl_es) 本文遵循"署名-非商业用途-保持一致&q ...

  7. 开源mp3播放器--madplay 编译和移植 简记

    madplay是一款开源的mp3播放器. http://madplay.sourcearchive.com/ 下面简单记录一下madplay的编译与移植到ARM开发板上的过程 一.编译x86版本的ma ...

  8. 利用json获取天气信息

    天气预报信息获取是利用json获取的,网上有非常多资源,源码.因为上面涉及到非常多天气信息,包含湿度,出行建议等,以及加入了全部城市代码的资源包.为了练手了解json的原理.我仅获取诚笃城市的最高温, ...

  9. [Android面试题-7] 写出一个Java的Singleton类(即单例类)

    1.首先明确单例的概念和特点: a>单例类只能有一个实例 b>单例类必须自己创建一个自己的唯一实例 c>单例类必须为其他所有对象提供这个实例 2.单例具有几种模式,最简单的两种分别是 ...

  10. GB2312引进和使用的字体

    一个:先上图看到的结果,下面的屏幕截图android在测试的结果"SD卡测试".."GPS测试"和其他字符24x24字体进来. 二:  1)简单介绍       ...