A1124 Raffle for Weibo Followers (20 point(s))

map充当散列

1. 原文

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...

2. 解析

抽奖 第一个为s ，第二个为s+n ，s+2n… 。领过了就s++；

用map记录领奖过的人

if (exist[str]>0){
   s++;
}

3. AC代码

#include<iostream>
#include<map>
#include<string>
using namespace std;
map<string,int> exist;
int main()
{
	int m,n,s;
	cin>>m>>n>>s;
	string str; int flag=0;
    for (int i = 1; i <= m; ++i)
    {
    	cin>>str;
    	if (exist[str]>0)
    	{
    		s++;
    	}
    	if (s==i&&exist[str]==0)
    	{
    		flag=1;
    		cout<<str<<endl;
    		s+=n;
    		exist[str]++;
    	}
    }
    if (flag==0)
    {
    	cout<<"Keep going..."<<endl;
    }
	return 0;
}