1124 Raffle for Weibo Followers (20 分)

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 (≤ 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...

 

简单标记即可

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n,m,k;
 4 string s[1005];
 5 map<string,int> mp;
 6 int main(){
 7     cin >> n >> m >> k;
 8     for(int i = 1 ; i <= n; i++){
 9         cin >> s[i];
10     }
11     if(k > n){
12         cout<<"Keep going..."<<endl;
13     }else{
14         while(k <= n){
15             if(mp[s[k]] == 0){
16                 mp[s[k]] = 1;
17                 cout << s[k] << endl;
18                 k += m;
19             }else{
20                 k++;
21             }
22         }
23     }
24     return 0;
25 }