大佬们大部分都用的map，我只用了string，比map麻烦一点。作为另一种思路，仅供参考。
题目如下：
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...参考代码如下：
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;
int main()
{
int n, k, st;
string s1[1010], s2; //s2用来储存出现过的id
cin >> n >> k >> st;
for (int i = 1; i <= n; i++)
cin >> s1[i];
if (st > n)
cout << "Keep going..."; //第一个中奖者st比总人数n大
else {
cout << s1[st] << endl;
s2 = s1[st]; //将第一个中奖者存到s2
for (int i = st + k; i <= n; i += k) {
if (s2.find(s1[i]) == string::npos) { //如果s1[i]没出现过
cout << s1[i] << endl;
s2 = s2 + " " + s1[i]; //将s1[i]添加到s2后面，记得中间加空格
}
else {
while (s2.find(s1[i]) != string::npos && i <= n) { //直到找到未出现的s1[i]
i++;
}
if (i <= n) { //如果最后i没超出范围
cout << s1[i] << endl;
s2 = s2 + " " + s1[i]; //将s1[i]添加到s2后面，记得中间加空格
}
}
}
}
system("pause");
return 0;
}