Nasty Hacks

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2299    Accepted Submission(s): 1816

Problem 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

解题思路:继续水题。直接比較每行的三个数。假设第一个数大于第二个数减去第三个数,则须要做广告。若等于。就无所谓。否则。不做广告。

AC代码:

#include <iostream>
#include <cstdio>
using namespace std; int main(){
// freopen("in.txt", "r", stdin);
int n, a, b, c;
while(scanf("%d", &n)==1){
for(int i=0; i<n; i++){
scanf("%d%d%d", &a, &b, &c);
if(a < b - c) printf("advertise\n");
else if(a == b - c) printf("does not matter\n");
else printf("do not advertise\n");
}
}
return 0;
}

HDU 2317 Nasty Hacks的更多相关文章

  1. 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 ...

  2. 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 ...

  3. Poj 3030 Nasty Hacks

    1.Link: http://poj.org/problem?id=3030 2.Content: Nasty Hacks Time Limit: 1000MS   Memory Limit: 655 ...

  4. 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 ...

  5. 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 ...

  6. hdu2317Nasty Hacks

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

  7. HOJ题目分类

    各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1 ...

  8. 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 ...

  9. OJ题解记录计划

    容错声明: ①题目选自https://acm.ecnu.edu.cn/,不再检查题目删改情况 ②所有代码仅代表个人AC提交,不保证解法无误 E0001  A+B Problem First AC: 2 ...

随机推荐

  1. 日志logging

    日志: 日志分为5个级别:debug(10),info(20),warning(30),error(40),critical(50) 日志四个组成部分:logger,handler,filter,fo ...

  2. teatime、

    Python之路,Day7 - 面向对象编程进阶   本节内容: 面向对象高级语法部分 经典类vs新式类 静态方法.类方法.属性方法 类的特殊方法 反射 异常处理 Socket开发基础 作业:开发一个 ...

  3. LeetCode(153) Find Minimum in Rotated Sorted Array

    题目 Total Accepted: 65121 Total Submissions: 190974 Difficulty: Medium Suppose a sorted array is rota ...

  4. DFS:POJ3620-Avoid The Lakes(求最基本的联通块)

    Avoid The Lakes Time Limit: 1000MS Memory Limit: 65536K Description Farmer John's farm was flooded i ...

  5. visual studio 的生成、重新生成、清理功能的说明

    生成 生成当前选中的项目,依赖的项目如果已经生成dll,则不生成,直接拷贝过来 重新生成 生成当前选中的项目,依赖的项目也会生成 清理 清除掉生成的dll和相关文件

  6. 【04】在webstorm里Export declarations are not supported by current JavaScript version

    [04]在webstorm里Export declarations are not supported by current JavaScript version     Export declara ...

  7. float.h

    float.h 一背景知识 浮点算术非常复杂   很多小的处理器在硬件指令方面甚至不支持浮点算术   其他的则需要一个独立的协处理器来处理这种运算   只有最复杂的计算机才在硬件指令集中支持浮点运算 ...

  8. 源码分析 脱壳神器ZjDroid工作原理

    0. 神器ZjDroid Xposed框架的另外一个功能就是实现应用的简单脱壳,其实说是Xposed的作用其实也不是,主要是模块编写的好就可以了,主要是利用Xposed的牛逼Hook技术实现的,下面就 ...

  9. 【bzoj2733】[HNOI2012]永无乡 线段树合并

    Description 永无乡包含 n 座岛,编号从 1 到 n,每座岛都有自己的独一无二的重要度,按照重要度可 以将这 n 座岛排名,名次用 1 到 n 来表示.某些岛之间由巨大的桥连接,通过桥可以 ...

  10. 关于时区、时间戳引起的bug理解

    时间戳定义:0时区1970年1月1日到现在的毫秒数,所以全世界同一时刻的时间戳都是一样的. 北京时间对应时间戳=unix(0时区对应时间的时间戳)-8*60*60*1000(8小时的毫秒数)----- ...