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...

已知M，N，S

M是人数，从S个人开始，每隔N个人进行抽取（抽到重复的就抽下一个），打印这些人。

如果抽取0人，则打印Keep going...

#include <iostream>
#include <map>
using namespace std;
int M, N, S;
char arr[99999][25];
map<string, bool> m;
int main() {
    scanf("%d%d%d", &M, &N, &S);
    for(int i = 1; i <= M; i++) 
        scanf("%s", arr[i]);
    for(int i = S; i <= M; i += N) {
        while(i <= M && m[arr[i]] == 1) i++;
        if(i > M) break;
        printf("%s
", arr[i]);
        m[arr[i]] = 1;
    }
    if(m.size() == 0) printf("Keep going...
");
    return 0;
}