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, Yue

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...

思路

1.map模拟一个字典dic记录该用户是否领过奖。
2.用一个bool值hasWinner来标记是否有人获过奖。 代码
#include<iostream>
#include<vector>
#include<map>
using namespace std;
int main()
{
int M,N,S;
while(cin >> M >> N >> S)
{
vector<string> List(M + );
map<string,int> dic;
bool hasWinner = false;
for(int i = ;i <= M;i++)
{
cin >> List[i];
} for(int i = S;i <= M;i += N)
{
while(dic.count(List[i]) > && i <= M) i++;
if(i > M) break;
hasWinner = true;
cout << List[i] << endl;
dic[List[i]]++;
}
if(!hasWinner)
cout << "Keep going..." << endl;
}
}

PAT1124:Raffle for Weibo Followers的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 1124 Raffle for Weibo Followers[简单]

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

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. RedHat系列软件管理(第二版) --源码包安装

    RedHat系列软件管理 --源码包安装 源码包特点: 拥有广泛的平台支持性,可以装在所有的类UNIX操作系统上,不用考虑CPU架构. 灵活性,可以在安装过程中指定特有的选项. 定制度非常高,可以自己 ...

  2. "《算法导论》之‘线性表’":基于动态分配的数组的顺序表

    我们利用静态分配的数组来实现的顺序表的局限还是挺大的,主要在于它的容量是预先定好的,用户不能根据自己的需要来改变.如果为了后续用户能够自己调整顺序表的大小,动态地分配数组空间还是很有必要的.基于动态分 ...

  3. java时间操作

    这篇讲的也很专业:http://soft.zdnet.com.cn/software_zone/2007/1129/660028.shtml java中的时间操作不外乎这四种情况: 1.获取当前时间 ...

  4. linux设备驱动模块引用和依赖

    /modules/04 # lsmod test 787 0 - Live 0xbf010000 (PO) func 633 1 test, Live 0xbf00c000 (PO) test -&g ...

  5. Java反编译工具(Java Decompiler)

    Java Decompiler是一种非常实用的JAVA反编译工具,可以对整个jar包进行反编译,也可以将其集成到eclipse上,非常方便的根据class文件的源码.,官网地址http://jd.be ...

  6. ES6之let命令

    ES6新增了let命令,用来声明变量.它的用法类似于var. let和var声明变量的区别: 1.let声明的变量,只在let命令所在的代码块内有效,出了这个块级作用域就不起作用 先看一个例子: { ...

  7. Tihinkphp3.2整合最新版阿里大鱼进行短信验证码发送

    阿里大鱼最新下载地址:阿里大鱼SDK下载 或者从官网进行下载:阿里大鱼SDK官网下载 下载完成后,将压缩包内的api_sdk文件夹放到ThinkPHP\Library\Vendor目录下,修改文件名为 ...

  8. x&(x-1)

    x&(x-1)可以用来求出x是否为2幂次方数:当&的结果为0时,x原值是2幂次方数,否则就不是2幂次方数: x=x&(x-1)即把x从低位开始的第一个1改成0.如1000,把1 ...

  9. 东方国信 - 软件开发人员面试问卷(ver1.001.002)

    1.    通用编程知识问卷(所有编程人员必做)... 1 1.1      SQL问卷... 1 1.2      翻译... 2 2.    Java问卷(Java程序员应答,其他跳过)... 2 ...

  10. jquery 设置占位符

    <script type="text/javascript">    $(document).ready(function(){       $('.inputfiel ...