PAT甲级1124

1124. Raffle for Weibo Followers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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...

#include<iostream>
#include<map>
#include<vector>
#include<string>
using namespace std;
int M, N, S;
map<string, bool> msb;
vector<string> vs;
int main()
{
  scanf("%d%d%d", &M, &N, &S);
  vs.push_back(" ");
  string s;
  for (int i = 0; i < M; i++)
  {
    cin >> s;
    vs.push_back(s);
    msb[s] = false;
  }
  int i = S;
  if (vs.size() - 1 < S)
  {
    cout << "Keep going..." << endl; return 0;
  }
  while (i <= M)
  {
    if (msb[vs[i]] == false)
    {
      cout << vs[i] << endl;
      msb[vs[i]] = true;
      i += N;
    }
    else {
      i++;
    }
  }
  return 0;
}