PAT A1124 Raffle for Weibo Followers (20point(s))

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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...

解答

#include<iostream>
#include<unordered_map>
#include<string>

using namespace std;

int main()
{
	int num, interval, winnerId, count = 0;
	unordered_map<string, bool> winners;
	string id;
	cin >> num >> interval >> winnerId;
	for (int i = 0; i < num; i++) {
		cin >> id;
		count++;
		if (count == winnerId && winners.find(id) == winners.end()) {
			printf("%s\n", id.c_str());
			winnerId += interval;
			winners[id] = true;
		}
		else if (count == winnerId) {
			count--;
		}
	}
	if (winners.size() == 0) printf("Keep going...\n");
	return 0;
}