zoukankan      html  css  js  c++  java
  • POJ 3030

    Description

    POJ 3030 - Nasty Hacks - Dilthey - 404 Not Found

    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, re 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 ≤ re ≤ 106 and 0 ≤ c ≤ 106.

    输入三个整数,分别是不打广告时的收益、打广告时的收益以及打广告要付出多少钱。

    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.

    输出“打广告!干他丫的!” or “打个毛线广告!” or “打不打好像没什么区别……”

    Sample Input

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

    Sample Output

    advertise
    does not matter
    do not advertise

    Source

     
    水题。
     1 #include<stdio.h>
     2 int main()
     3 {
     4   int r,e,c,n;scanf("%d",&n);
     5   while(n){
     6     scanf("%d%d%d",&r,&e,&c);
     7     if(r>e-c) printf("do not advertise
    ");
     8     else if(r==e-c) printf("does not matter
    ");
     9     else printf("advertise
    ");
    10     n--;
    11   }
    12 }
    转载请注明出处:https://dilthey.cnblogs.com/
  • 相关阅读:
    python之路--day22--多态....property..calssmethod....staticmethod
    python之路--day21--组合与封装
    内存数据库
    什么时候调用:拷贝构造函数、赋值运算符
    类成员初始化顺序
    HTTP、TCP、UDP的区别
    进程和线程的区别
    C++中类的静态成员与实例成员的区别
    设计模式
    const函数
  • 原文地址:https://www.cnblogs.com/dilthey/p/6804195.html
Copyright © 2011-2022 走看看