1124 Raffle for Weibo Followers (20)

1124 Raffle for Weibo Followers (20)（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 (<= 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...

题目大意：微博抽奖，给出一串待抽奖人的名字， 给出第一个抽奖的序号s,给出抽奖序号之间的间隔n， 给出所有待抽奖人的人数m; 输出所有被抽中的人的名字， 如果没有一个人被抽中，输出特定字符串。
如果某一个人第二次被选中，则跳过这个人，继续向下抽奖；
思路：用set类型来保存抽中的姓名，可以有效的筛选以及选中的人

 1 #include<iostream>
 2 #include<set>
 3 using namespace std;
 4 int main(){
 5   int m, n, s, i;
 6   cin>>m>>n>>s;
 7   string person[1000];
 8   set<string> winner;
 9   bool flag = true;
10   for(i=0; i<m; i++) cin>>person[i];
11   for(i=s-1; i<m; i+=n){
12     if(winner.count(person[i])==0){
13       flag = false;
14       cout<<person[i]<<endl;
15       winner.insert(person[i]);
16     }else{
17       while(winner.count(person[i]) && i<m){i++;};
18       cout<<person[i]<<endl;
19     }
20   }
21   if(flag) cout<<"Keep going..."<<endl;
22   return 0;
23 }