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...

思路：i从第一个下标开始遍历，如果其对应的选手没有获胜，就间隔n继续遍历，如果获胜，就+1继续遍历，用map存储是否获胜

#include<bits/stdc++.h>
using namespace std;
const int maxn=10010;
string s[maxn];
map<string,bool> mp;

int main(){
    int m,n,t;
    cin>>m>>n>>t;
    for(int i=0;i<m;i++){
        cin>>s[i];
        mp[s[i]]=false;
    }
    if(m<t){
        cout<<"Keep going..."<<endl;
        return 0;
    }
    int k=0;
    for(int i=t-1;i<m; ){
        if(mp[s[i]]==false){
            cout<<s[i]<<endl;
            mp[s[i]]=true;
            i+=n;
        }
        else{
            i++;
        }
    }
    return 0;
}