1124 Raffle for Weibo Followers-PAT甲级

最新推荐文章于 2023-05-20 16:13:09 发布

星辰浩宇

最新推荐文章于 2023-05-20 16:13:09 发布

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分类专栏： pat甲级

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本文介绍了一个简单的抽奖算法实现，用于从微博转发中选出获奖者。算法使用C++编写，避免重复中奖并实时处理输入。

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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...

很简单的一道题，不需要用数组去处理，输入一个处理一个即可

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
int m,n,s;
map<string,int>mp;
vector<string>ve;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>m>>n>>s;
	int flag=0;
	for(int i=1;i<=m;i++){
		string t;;
		cin>>t;
		if(i==s){
			if(mp[t]==0){
				mp[t]=1;
				flag=1;
				ve.push_back(t);
				s+=n;
			}else{
				s++;
			}
		}
	}
	if(!flag) cout<<"Keep going..."<<endl;
	else{
		for(int i=0;i<ve.size();i++){
			cout<<ve[i]<<endl;
		}
	}
	return 0;
}