PAT甲级——A1124 Raffle for Weibo Followers

John got a full mark on PAT. He was so happy that he decided to hold a raffle（抽奖） for his followers on Weibo -- that is, he would select winners from every N followers who forwarded his post, and give away gifts. Now you are supposed to help him generate the list of winners.

Input Specification:

Each input file contains one test case. For each case, the first line gives three positive integers M ( $\leq 1000), N and S, being the total number of forwards, the skip number of winners, and the index of the first winner (the indices start from 1). Then M lines follow, each gives the nickname (a nonempty string of no more than 20 characters, with no white space or return) of a follower who has forwarded John's post.$

Note: it is possible that someone would forward more than once, but no one can win more than once. Hence if the current candidate of a winner has won before, we must skip him/her and consider the next one.

Output Specification:

For each case, print the list of winners in the same order as in the input, each nickname occupies a line. If there is no winner yet, print Keep going...instead.

Sample Input 1:

9 3 2
Imgonnawin!
PickMe
PickMeMeMeee
LookHere
Imgonnawin!
TryAgainAgain
TryAgainAgain
Imgonnawin!
TryAgainAgain

Sample Output 1:

PickMe
Imgonnawin!
TryAgainAgain

Sample Input 2:

2 3 5
Imgonnawin!
PickMe

Sample Output 2:

Keep going...

 1 #include <iostream>
 2 #include <unordered_map>
 3 #include <string>
 4 using namespace std;
 5 int m, k, s;
 6 int main()
 7 {
 8     cin >> m >> k >> s;
 9     unordered_map<string,int>res;
10     string str;
11     for (int i = 1; i <= m; ++i)
12     {
13         cin >> str;
14         if (i == s)
15         {
16             if (res[str] == 1)//输出过
17                 s++;//后移动
18             else
19             {
20                 cout << str << endl;
21                 res[str] = 1;
22                 s += k;
23             }
24         }
25     }
26     if (res.size() == 0)
27         cout << "Keep going..." << endl;
28     return 0;
29 }