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! TryAgainAgainSample Output 1:
PickMe Imgonnawin! TryAgainAgainSample Input 2:
2 3 5 Imgonnawin! PickMeSample Output 2:
Keep going...
题目思路：
一个字符串数组，输出固定间隔的字符串，如果当前字符串已经输出过一次，那么输出下一个字符串。我们可以采用map的key记录字符串，value记录字符串被抽中的次数。
题目代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;int m, n, s;
string t[1001];
//用来记录被抽中的次数
map<string,int> vis;int main()
{cin>>m>>n>>s;int cnt = 0;for(int i=0 ;i<m ;i++){cin>>t[i];vis[t[i]] = 0;}//flag用来标记是否有中奖者 int flag = 1;//i=s-1 直接从中奖者开始遍历 for(int i=s-1; i<m; ){if(vis[t[i]]==0){flag = 0;cout<<t[i]<<endl;vis[t[i]] = 1;//如果有中奖者 直接+n都下一个中奖者 i += n;}else{//如果该中奖者已经中过一次 则跳到下一个人 i += 1;}}if(flag){puts("Keep going...");}return 0;
}