POJ3030——Nasty Hacks

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#POJ3030 #Nasty Hacks #广告

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作为NastyHacks公司的CEO，您需要决定是否通过广告来增加销售额。分析师提供了广告投放前后预计的营收数据。根据这些数据，需要制定出最有利的广告策略。

Nasty Hacks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13634		Accepted: 9460

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

Nordic 2006

解：

根据不做广告时的预期收入 r，做广告时的预期收入 e和广告的成本 c来判断是否做广告或者不重要，即判断 r和 e-c的大小关系即可。

#include<stdio.h>
  
int main()  
{  
    int t,r,e,c;  
    while(scanf("%d",&t)!=EOF)  
    {  
        while(t--)  
        {  
            scanf("%d%d%d",&r,&e,&c);  
            if(r>(e-c))
				printf("do not advertise\n");  
            else if(r<(e-c))
				printf("advertise\n");  
            else 
				printf("does not matter\n");  
        }  
    }  
    return 0;  
}