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...
 #include <iostream>
#include <unordered_map>
#include <string>
using namespace std;
int m, k, s;
int main()
{
cin >> m >> k >> s;
unordered_map<string,int>res;
string str;
for (int i = ; i <= m; ++i)
{
cin >> str;
if (i == s)
{
if (res[str] == )//输出过
s++;//后移动
else
{
cout << str << endl;
res[str] = ;
s += k;
}
}
}
if (res.size() == )
cout << "Keep going..." << endl;
return ;
}

PAT甲级——A1124 Raffle for Weibo Followers的更多相关文章

  1. PAT甲级 1124. Raffle for Weibo Followers (20)

    1124. Raffle for Weibo Followers (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN ...

  2. PAT A1124 Raffle for Weibo Followers (20 分)——数学题

    John got a full mark on PAT. He was so happy that he decided to hold a raffle(抽奖) for his followers ...

  3. A1124. Raffle for Weibo Followers

    John got a full mark on PAT. He was so happy that he decided to hold a raffle(抽奖) for his followers ...

  4. PAT_A1124#Raffle for Weibo Followers

    Source: PAT A1124 Raffle for Weibo Followers (20 分) Description: John got a full mark on PAT. He was ...

  5. PAT甲级:1124 Raffle for Weibo Followers (20分)

    PAT甲级:1124 Raffle for Weibo Followers (20分) 题干 John got a full mark on PAT. He was so happy that he ...

  6. PAT 1124 Raffle for Weibo Followers

    1124 Raffle for Weibo Followers (20 分)   John got a full mark on PAT. He was so happy that he decide ...

  7. pat 1124 Raffle for Weibo Followers(20 分)

    1124 Raffle for Weibo Followers(20 分) John got a full mark on PAT. He was so happy that he decided t ...

  8. 1124 Raffle for Weibo Followers (20 分)

    1124 Raffle for Weibo Followers (20 分) John got a full mark on PAT. He was so happy that he decided ...

  9. PAT1124:Raffle for Weibo Followers

    1124. Raffle for Weibo Followers (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN ...

随机推荐

  1. Linkedlist 详解

    基本介绍 Linkedlist基于链表的动态数组(双向链表): 可以被当作堆栈(后进先出).队列(先进先出)或双端队列进行操作. 数据添加删除效率高,只需要改变指针指向即可,但是访问数据的平均效率低, ...

  2. TopCoder[SRM587 DIV 1]:TriangleXor(550)

    Problem Statement      You are given an int W. There is a rectangle in the XY-plane with corners at ...

  3. 01二维背包——poj2576

    /* 要求把a数组分成两个集合,两个集合人数最多差1,并且元素之和的差尽可能小 那只要把所有可行的列出来即可 01二维背包,即体积是个二维数据,那么我们的背包状态也应该设为二维 dp[j][k]设为 ...

  4. Nginx被动健康检查和主动健康检查

    1.被动健康检查 Nginx自带有健康检查模块:ngx_http_upstream_module,可以做到基本的健康检查,配置如下: upstream cluster{ server max_fail ...

  5. delphi windows操作

    输入 procedure TypeKeyString(s: string); var c: Char; i: integer; off: integer; vkw: Word; begin to Le ...

  6. NX文件名与工程图名自动关联

    1.先去D:\Program Files\Siemens\NX 9.0\LOCALIZATION\prc\simpl_chinese\startup里,把默认的图框模板替换成自己定制好的模板,如何替换 ...

  7. hdu多校第五场1006 (hdu6629) string matching Ex-KMP

    题意: 给你一个暴力匹配字符串公共前缀后缀的程序,为你对于某个字符串,暴力匹配的次数是多少. 题解: 使用扩展kmp构造extend数组,在扩展kmp中,设原串S和模式串T. extend[i]表示T ...

  8. hdu4126_hdu4756_求最小生成树的最佳替换边_Kruskal and Prim

    目录 Catalog Solution: (有任何问题欢迎留言或私聊 && 欢迎交流讨论哦 Catalog Problem:  Portal: hdu4126 hdu4756  原题目 ...

  9. FZU - 2295 Human life:网络流-最大权闭合子图-二进制优化-第九届福建省大学生程序设计竞赛

    目录 Catalog Solution: (有任何问题欢迎留言或私聊 && 欢迎交流讨论哦 http://acm.fzu.edu.cn/problem.php?pid=2295 htt ...

  10. opencv-图像形态学之开运算、闭运算、形态学梯度、顶帽、黑帽合辑

    转自:https://blog.csdn.net/poem_qianmo/article/details/24599073 1.1 开运算(Opening Operation) 开运算(Opening ...