PAT（甲级）2017年春季考试 PAT 1124 Raffle for Weibo Followers (20 分)

PAT（甲级）2017年春季考试
PAT 1124 Raffle for Weibo Followers (20 分)

【题目描述】
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...

【解题思路】

微博抽奖，输入给定第一个中奖用户编号，以及后面每隔多少个用户给一个奖。

用一个循环输入，记录当前是第几条记录，判断当前用户是否中奖，输出即可。

需要注意的是，已经中奖的用户不再重复中奖，我们使用一个set，将中奖用户加入，如果后面遇到相同的字符串则跳过。

【满分代码】

#include <iostream>
#include <cstdio>
#include <set>
using namespace std;
int main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int m,n,s,temp=0;
	string t;
	set<string> st;
	cin>>m>>n>>s;
	while(m--)
	{
		cin>>t;
		if(st.count(t)==0)
			temp++;
		else continue;//没加上这句也能过，测试数据有漏洞
		if(temp-s>=0&&(temp-s)%n==0)
		{
			cout<<t<<endl;
			st.insert(t);
		}
	}
	if(st.size()==0)
		cout<<"Keep going..."<<endl;
	return 0;
}