PAT 1124 Raffle for Weibo Followers

最新推荐文章于 2023-09-30 15:57:15 发布

原创最新推荐文章于 2023-09-30 15:57:15 发布 · 111 阅读

0 ·

CC 4.0 BY-SA版权

刷题专栏收录该内容

94 篇文章

订阅专栏

约翰在微博上举办抽奖活动，从转发他帖子的每N名粉丝中选出赢家，这份包含所有获奖者昵称的完整名单揭示了他们的幸运时刻。

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.
在这里插入图片描述
题目大意：
确定第一个后依次确定后面的获奖者，如果该获奖者以及获过奖，就跳到下一个。
思路：

用一个map来记录获奖者名单，值为1，如果之前获过奖，就跳过，到下一个。
关于跳过到下一个这里，不能使用纯粹输入时的i来计数，应该还是需要一个量来记录，遇到第一个获奖者就变成0，开始计数，如果到下一个获奖者是已获的，则保持不变，继续遍历，直到找到下一个获奖者。
代码：

#include<iostream>
#include<vector>
using namespace std;
vector<string> v;
bool check(string s){
	bool flag=false;
	for(int i=0;i<v.size();i++) {
		if(v[i]==s) flag=true;
	}
	return flag;
}
int main(){
	int m,n,s;
	cin>>m>>n>>s;
	int snum=0,nnum=0;
	while(m--){
		string str;
		cin>>str;
		snum++;
		if(snum==s){
			v.push_back(str);
	    	continue;
		} 
		else if(snum>s){
			if(nnum+1==n){
				if(check(str)) continue;
				else{
					v.push_back(str);
					nnum=0;
				} 
			} 
			else nnum++;
		}
	} 
	if(v.size()==0) printf("Keep going...");
	else{
		for(int i=0;i<v.size();i++) cout<<v[i]<<endl;
	}
	return 0;
}