1.Link:

http://poj.org/problem?id=3030

2.Content:

Nasty Hacks
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 12350   Accepted: 8537

Description

You are the CEO of Nasty Hacks Inc., a company that creates small pieces of malicious software which teenagers may use to fool their friends. The company has just finished their first product and it is time to sell it. You want to make as much money as possible and consider advertising in order to increase sales. You get an analyst to predict the expected revenue, both with and without advertising. You now want to make a decision as to whether you should advertise or not, given the expected revenues.

Input

The input consists of n cases, and the first line consists of one positive integer giving n. The next n lines each contain 3 integers, r, e and c. The first, r, is the expected revenue if you do not advertise, the second, e, is the expected revenue if you do advertise, and the third, c, is the cost of advertising. You can assume that the input will follow these restrictions: −106 ≤ r, e ≤ 106 and 0 ≤ c ≤ 106.

Output

Output one line for each test case: “advertise”, “do not advertise” or “does not matter”, presenting whether it is most profitable to advertise or not, or whether it does not make any difference.

Sample Input

3
0 100 70
100 130 30
-100 -70 40

Sample Output

advertise
does not matter
do not advertise

Source

3.Method:

4.Code:

 #include<iostream>
using namespace std;
int main()
{
int i,n;
int r,e,c;
int result;
cin>>n;
for(i=;i<n;i++)
{
cin>>r>>e>>c;
result=e-c-r;
if(result>) cout<<"advertise"<<endl;
else if(result<) cout<<"do not advertise"<<endl;
else cout<<"does not matter"<<endl;
}
//system("pause");
return ; }

5.Reference:

Poj 3030 Nasty Hacks的更多相关文章

  1. HDOJ 2317. Nasty Hacks 模拟水题

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tota ...

  2. HDOJ(HDU) 2317 Nasty Hacks(比较、)

    Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of mali ...

  3. Nasty Hacks <入门练手题>

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total S ...

  4. HDU 2317 Nasty Hacks

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tota ...

  5. 【转】POJ百道水题列表

    以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight ...

  6. POJ解题经验交流

    感谢范意凯.陈申奥.庞可.杭业晟.王飞飏.周俊豪.沈逸轩等同学的收集整理.   题号:1003 Hangover求1/2+1/3+...1/n的和,问需多少项的和能超过给定的值 类似于Zerojudg ...

  7. 算法之路 level 01 problem set

    2992.357000 1000 A+B Problem1214.840000 1002 487-32791070.603000 1004 Financial Management880.192000 ...

  8. hdu2317Nasty Hacks

    Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of mali ...

  9. dir命令只显示文件名

    dir /b 就是ls -f的效果 1057 -- FILE MAPPING_web_archive.7z 2007 多校模拟 - Google Search_web_archive.7z 2083 ...

随机推荐

  1. 理解WebKit和Chromium: 调试Android系统上的Chromium

    转载请注明原文地址:http://blog.csdn.net/milado_nju 1. Android上的调试技术 在Android系统上,开发人员能够使用两种不同的语言来开发应用程序,一种是Jav ...

  2. iOS开发——实用篇&KVO与KVC详解

    KVO与KVC详解 由于ObjC主要基于Smalltalk进行设计,因此它有很多类似于Ruby.Python的动态特性,例如动态类型.动态加载.动态绑定等.今天我们着重介绍ObjC中的键值编码(KVC ...

  3. cocos2d粒子效果

    第9章 粒子效果 游戏开发者通常使用粒子系统来制作视觉特效.粒子系统能够发射大量细小的粒子并对他们进行渲染,而且效率要远高于渲染同样数目的精灵.粒子系统可以模拟下雨.火焰.雪.爆炸.蒸气拖尾以及其他多 ...

  4. debian7 oracle11g 解决 link binaries 错误方案

    ln -s /etc /etc/rc.d ln -s /usr/bin/awk /bin/awk ln -s /usr/bin/basename /bin/basename ln -s /usr/bi ...

  5. 新一代 PHP 加速插件 Zend Opcache

    参考:http://www.laogui.com/Zend-Opcache 大家知道目前PHP的缓存插件一般有三个:APC.eAccelerator.XCache,但未来它们可能都会消失,因为PHP ...

  6. MPEG简介 + 如何计算CBR 和VBR的MP3的播放时间

    1. 声明本文所写内容,多数整理自互联网,版权归原作者所有笔者知识有限,文中难免有误,欢迎批评指正,admin (at) crifan.com觉得此文对你有帮助,想要发邮件来感谢的,也欢迎哈,^_^欢 ...

  7. CSS 实现行内和上下自适应的几种方法

    在写一个移动端网页,发现网页的头部搜索框两边各有固定宽度的按钮,搜索框可以根据宽度的变化来改变自己的宽度,达到填充的目的,也就是一种自适应吧,下面写写自己尝试的几种方法 一 利用css3 的width ...

  8. cmd运行java,含传参,引用jar

    1,创建一个java project,完成编码 在Eclipse的资源管理器中选中你要打包的项目,右键点击,选择“导出”项,弹出导出对话框,在下面的Java目录下选择“JAR 文件”项,下一步,在导出 ...

  9. Share_memory

    共享内存是允许多个进程共享一块内存,由此来达到交换信息的进程通信机制:它很快没有中间介质,唯一的不足就是需要一定的同步机制控制多个进程对同一块内存的读/写,,它的原理如下: 每个共享内存段都有一个sh ...

  10. Linux下RPM软件包的安装及卸载

    http://os.51cto.com/art/201001/177866.htm 在 Linux 操作系统下,几乎所有的软件均通过RPM 进行安装.卸载及管理等操作.RPM 的全称为Redhat P ...