PAT甲级——1124 Raffle for Weibo Followers （set或者unordered_set的应用)

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...

题目大意：在转发转发微博的人（一堆名字）中按照设定好的规则选择中奖者；M是转发的人数，N是选择winner时间隔的人数，S是第一个winner的index。。。

思路：把名字依次放入字符串数组name中，遍历name，若当前的转发者不在集合se中，则为winner，输出人名并且将它放入se中；若当前的转发者在se中，则往后找name；遍历name之后如果se为空，则输出“Keep going...”

#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
using namespace std;

int main()
{
	int M, N, S;
	scanf("%d%d%d", &M, &N, &S);
	unordered_set<string> se;
	vector<string> name;
	for (int i = 0; i < M; i++) {
		string tmp;
		cin >> tmp;
		name.push_back(tmp);
	}
	for (int i = S-1; i < name.size(); i = i + N) {
		while (1) {
			if (i >= name.size())
				break;
			if (se.find(name[i]) == se.end())
				break;
			i++;
		}
		if (i < name.size()) {
			cout << name[i] << endl;
			se.insert(name[i]);
		}
	}
	if (se.empty())
		printf("Keep going...\n");
}