1. 題目

John got a full mark on PAT. He was so happy that he decided to hold a raffle（抽獎） 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...

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...

2. 題意

John進行轉發抽獎活動，給出三位正整數M，N，S，分別表示轉發的總人數、小明決定的中獎間隔、以及第一名中獎者的序號（編號從1開始）。每位網友可能多次轉發，但只能中一次獎。如果處於當前位置的網友中過獎，則一直順延到下一位沒有中獎的網友。

3. 思路——模擬+map

1. 如果第1位中獎者的序號大於轉發總人數，說明沒人中獎，直接輸出return 0即可。
2. 否則從第1位中獎者S開始，每N位產生一名中獎者：如果該名網友先前中獎過，那麼就順延給下一位沒有中獎的網友；否則就輸出該名中獎者的網名，並將其置為中獎過的狀態！

• 錯誤總結：
  • 這題我自己因為沒有理解對題目意思，所以卡了很久，剛開始題目有點沒看懂，這是硬傷，後面看了翻譯，又理解錯了題目，以為順延值順延一位，沒有想到順延那一位如果也中獎過再往下順延的情況。綜上理解，導致卡了很久，題目本身並不難。
  • 還有就是剛開始想到用set來判斷網友有沒有中獎過，但是這比起使用map麻煩很多，所以選擇正確的資料結構很有必要！

4. 程式碼

#include <iostream>
#include <map>
#include <string>

using namespace std;

int main()
{
	int m, n, s;
	cin >> m >> n >> s;
	map<string, int> res;
	string names[m + 1];
	for (int i = 1; i <= m; ++i) 
		cin >> names[i];
		
	if (s > m) // 如果第一位獲勝者的位數大於轉發總數，說明沒有獲勝者 
		cout << "Keep going..." << endl;

	// 從第一位獲勝者s開始，每間隔n位產生一名獲勝者 
	for (int i = s; i <= m; i += n)
	{
		if (res[names[i]] == 1) // 如果這位獲勝者已經拿過一次獎了 
		{ 
			i++;	// 那麼開始順延給下一位 
			while (i <= m)	// 一直順延，直到沒有獲獎過之前都沒獲獎過的那一位 
			{
				if (res[names[i]] == 0)  
				{
					cout << names[i] << endl;
					res[names[i]] = 1; 
					break;
				}
				i++;
			}
		} else	// 如果這位獲勝者未獲過獎，則輸出 
		{
			cout << names[i] << endl;
			res[names[i]] = 1;
		}
	
	}
	return 0;
}