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 = 1; i <= M; i++){
        cin >> ss;
        if(i == next){
            if(mp.count(ss) == 0){
                ans.push_back(ss);
                mp[ss] = i;
                next = next + N;
            }else{
                next++;
            }
        }
    }
    if(ans.size() == 0)
        cout << "Keep going..." << endl;
    else{
        for(int i = 0; i < ans.size(); i++)
            cout << ans[i] << endl;
    }
    cin >> N;
    return 0;
}