PAT 1124 Raffle for Weibo Followers

1124 Raffle for Weibo Followers (20 分)

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...

#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+1);
    for(int i=1;i <= m;i++){
        string t;cin >> t;
        vec[i] = t;
    }
    if(s > m) {printf("Keep going...
");return 0;}


    map<string,int> mp;
    for(int i=s;i <= m;){
        if(mp[vec[i]] == 0){
            cout << vec[i] << endl;
            mp[vec[i]] = 1;
            i += n;
        }
        else{
            i++;
        }
    }



    return 0;
}

发现一个性质，题面越长一般越简单。。