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<bits/stdc++.h>

using namespace std;

typedef long long ll;

int main(){

    int m,n,s;

    cin >> m >> n >> s;

    vector<string> vec(m+);

    for(int i=;i <= m;i++){

        string t;cin >> t;

        vec[i] = t;

    }

    if(s > m) {printf("Keep going...\n");return ;}

    map<string,int> mp;

    for(int i=s;i <= m;){

        if(mp[vec[i]] == ){

            cout << vec[i] << endl;

            mp[vec[i]] = ;

            i += n;

        }

        else{

            i++;

        }

    }

    return ;

}

发现一个性质，题面越长一般越简单。。