HDU2317 Nasty Hacks【水题】

Nasty Hacks

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

Total Submission(s): 2620    Accepted Submission(s): 2047

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

Nordic 2006

题目大意：公司销售商品的时候在考虑做不做广告，给你不做广告的预期效益、做广告的预期

效益和做广告的花销。为了效益问题，问：是否该做广告。做广告就输出"advertise"，不做广

告就输出"do not advertise"，做不做都一样就输出"does not matter"。

思路：计算不做广告的效益和左广告的效益-做广告的花销哪个高，然后根据情况输出不同结果。

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;int main(){    int T;    cin >> T;    while(T--)    {        int r,e,c;        cin >> r >> e >> c;        if(r > e - c)            cout << "do not advertise" << endl;        else if(r == e - c)            cout << "does not matter" << endl;        else            cout << "advertise" << endl;    }    return 0;}