PAT (Advanced Level) -1124 Raffle for Weibo Followers_1124 raffle for weibo followers测试点2-优快云博客

博客围绕微博转发抽奖问题展开，介绍了PAT基础级别的相关题目。给出输入规范，包含转发总数、中奖间隔数和首个中奖者索引等信息，还说明了输出规范。同时提供了两组示例输入及对应的输出，帮助理解抽奖逻辑。

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1124 Raffle for Weibo Followers（20 分）

PAT (Basic Level) -1069 微博转发抽奖（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...

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

using namespace std;
map<string,int>mp;

int main() 
{
	int m,n,s;
	string name[1002];
	cin>>m>>n>>s;
	for(int i=1;i<=m;i++)
	{
		cin>>name[i];
	}
	if(s>m)
	{
		cout << "Keep going..." << endl;
	}
	else
	{
		cout<<name[s]<<endl;
		mp[name[s]]++;
		int j=s+n;
		while(j<=m)
		{
			if(mp[name[j]]==0)
			{
				mp[name[j]]++;
				cout<<name[j]<<endl;
				j+=n;
			}
			else j++;
		}	
	}
		
}