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 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 <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <set>
using namespace std;
const int MAX = 1e3 + ; int m, n, s, cnt = ;
struct node
{
char s[];
}P[MAX], S[MAX];
set <string> st;
pair <set <string> :: iterator, bool> pr;
set <string> :: iterator iter; int main()
{
// freopen("Date1.txt", "r", stdin);
scanf("%d%d%d", &m, &n, &s);
for (int i = ; i <= m; ++ i)
scanf("%s", &P[i].s); for (int i = s; i <= m; i += n)
{
pr = st.insert(P[i].s);
if (pr.second)
{
strcpy(S[cnt ++].s, P[i].s);
continue;
}
while (i <= m)
{
++ i;
pr = st.insert(P[i].s);
if (pr.second)
{
strcpy(S[cnt ++].s, P[i].s);
break;
}
}
} if (cnt == )
{
printf("Keep going...\n");
return ;
}
for (int i = ; i < cnt; ++ i)
printf("%s\n", S[i].s);
return ;
}
pat 1124 Raffle for Weibo Followers(20 分)的更多相关文章
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 1124 Raffle for Weibo Followers
题意:水题,直接贴代码了.(为什么我第一遍做的时候代码写的那么烦?) 代码: #include <iostream> #include <string> #include &l ...
- 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 ...
- 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
- PHP 调试脚本
如果想要通过php.exe直接运行和调试脚本,可以在PHPStorm配置如下: 1.PHP安装XDebug的扩展. 2.在PHPStorm中,配置XDebug: 1) 打开菜单 "文件&qu ...
- 【Leetcode 做题学算法周刊】第一期
首发于微信公众号<前端成长记>,写于 2019.10.28 背景 本文记录刷题过程中的整个思考过程,以供参考.主要内容涵盖: 题目分析设想 编写代码验证 查阅他人解法 思考总结 目录 1. ...
- 百万年薪python之路 -- 小数据池和代码块练习
1.请用代码验证 "alex" 是否在字典的值中? info = {'name':'王刚蛋','hobby':'铁锤','age':'18',...100个键值对} info = ...
- 通过FeignClient接收shaded的javabean的JSON序列化
问题说明 最近做了关于flink的需求. 现在需要通过HTTP访问FLINK的 RESTAPI, rest 接口的JSON 非常庞大而复杂. 那么怎么去完整的接收数据呢? 方法一就是手写部分需要的Ja ...
- js奥义:原型与原型链(1)
要弄懂原型链,首先应先明白prototype原型对象.__proto__.对象三者之间的关系. 引入构造函数的相关定义: 构造函数是一种比较特殊的函数,用于批量实例化对象.通俗一点说,构造函数是用于生 ...
- 利用X-Forwarded-For伪造客户端IP漏洞成因及防范
内容转载自叉叉哥https://blog.csdn.net/xiao__gui/article/details/83054462 问题背景 在Web应用开发中,经常会需要获取客户端IP地址.一个典型的 ...
- 《JavaScript设计模式与开发实践》-- 代理模式
详情个人博客:https://shengchangwei.github.io/js-shejimoshi-daili/ 代理模式 1.定义 代理模式:代理模式是为一个对象提供一个代用品或占位符,以便控 ...
- 【翻译】Prometheus 2.12.0 新特性
Prometheus 2.12.0 现在(2019.08.17)已经发布,在上个月的 2.11.0 之后又进行了一些修正和改进. 在当前的 6 周发布周期中,每一个 Prometheus 版本都有比较 ...
- Appium的加载过程
appium运行流程 Appium的加载过程如上图. 1)调用Android adb完成基本的系统操作: 2)向Android上部署bootstrap.jar: 3)Bootstrap.jar For ...
- NOIP模拟 22
剧情回放:xuefeng:考场上你们只打暴力不打正解,我不满意! skyh:考场怒切T2以表明自己拥护xuefeng的决心 BoboTeacher:这场考试就没想让你们上100 神犇skyh:(笑而不 ...