ACM STEP 1.2.3 Nasty Hacks
来源:互联网 发布:python xml dom 编辑:程序博客网 时间:2024/04/29 19:17
Nasty Hacks
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2161 Accepted Submission(s): 1359Problem 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.
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
30 100 70100 130 30-100 -70 40
Sample Output
advertisedoes not matterdo not advertise
Source
Nordic 2006
Recommend
zty
#include <iostream>using namespace std;int main(){ double r,e,c,m,s; int i; void display(int t); cin >>i; while(i--) { cin >>r; cin >>e; cin >>c; m=e-c; if(m>r) s=1; else if(m==r) s=2; else s=3; display(s); } return 0;}void display(int t){ switch (t) { case 1: cout <<"advertise"<<endl; break; case 2: cout <<"does not matter"<<endl; break; case 3: cout <<"do not advertise"<<endl; break; default: break; }}
- ACM STEP 1.2.3 Nasty Hacks
- Nasty Hacks
- Nasty Hacks
- POJ 3030 Nasty Hacks
- poj 3030 Nasty Hacks
- 2317:Nasty Hacks
- HDU 2317 Nasty Hacks
- HOJ 2453 Nasty Hacks
- poj 3030 Nasty Hacks
- HDU2317:Nasty Hacks
- hdu-2317-Nasty Hacks
- Nasty Hacks 2317
- hd 2317 Nasty Hacks
- hdu1.2.3 Nasty Hacks
- Nasty Hacks
- HDU 2317 Nasty Hacks
- Poj 3030 Nasty Hacks
- HDU-2317-Nasty Hacks
- 共用体和枚举类型
- Visitor Pattern
- Mysql的死锁问题,解决Locked状态
- 初学hadoop之--------java中的for each语句----for (IntWritable val : values)
- 详解SQL Server Profiler分析死锁几大步骤
- ACM STEP 1.2.3 Nasty Hacks
- 我的新CSDN
- linux 内存释放 (转)
- Matlab画图入门
- GDI+初探之图像透明显示和保存
- ACM STEP 1.2.4 find your present (2)
- 网络图片的获取和显示的方法
- eCos中的main函数
- 2011成都网选最水的几道题