1124 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...

题意：

　　给出粉丝列表，根据第一个粉丝的位置，以及每两个粉丝之间的间距，来寻找满足要求的所有粉丝（每个粉丝只能获奖一次，如果重复获奖的话则自动寻找下一个，知道找到没有获奖经历的那个，或者到达粉丝列表的末尾）

思路：

　　用set存储已经获过奖的人的姓名。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int m, n, k, index;
    cin >> m >> n >> k;
    set<string> s;
    vector<string> followers(m + 1), ans;
    for (int i = 1; i <= m; ++i) {
        cin >> followers[i];
    }
    if (m < k)
        ans.push_back("Keep going...");
    else {
        while (k <= m) {
            if (s.find(followers[k]) == s.end()) {
                s.insert(followers[k]);
                ans.push_back(followers[k]);
            } else {
                while (k <= m && s.find(followers[k]) != s.end()) {
                    k++;
                }
                ans.push_back(followers[k]);
            }
            k += n;
        }
    }
    for (int i = 0; i < ans.size(); ++i) {
        cout << ans[i] << endl;
    }

    return 0;
}