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 (<= 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...

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<map>
 #include<string>
 using namespace std;
 map<string, int> mp;
 vector<string> ans;
 int main(){
     int M, N, S;
     string ss;
     scanf("%d%d%d", &M, &N, &S);
     int next = S;
     for(int i = ; i <= M; i++){
         cin >> ss;
         if(i == next){
             if(mp.count(ss) == ){
                 ans.push_back(ss);
                 mp[ss] = i;
                 next = next + N;
             }else{
                 next++;
             }
         }
     }
     if(ans.size() == )
         cout << "Keep going..." << endl;
     else{
         for(int i = ; i < ans.size(); i++)
             cout << ans[i] << endl;
     }
     cin >> N;
     return ;
 }