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...
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 #include <map> 6 #include <stack> 7 #include <vector> 8 #include <queue> 9 #include <set> 10 using namespace std; 11 const int MAX = 1e3 + 10; 12 13 int m, n, s, cnt = 0; 14 struct node 15 { 16 char s[30]; 17 }P[MAX], S[MAX]; 18 set <string> st; 19 pair <set <string> :: iterator, bool> pr; 20 set <string> :: iterator iter; 21 22 int main() 23 { 24 // freopen("Date1.txt", "r", stdin); 25 scanf("%d%d%d", &m, &n, &s); 26 for (int i = 1; i <= m; ++ i) 27 scanf("%s", &P[i].s); 28 29 for (int i = s; i <= m; i += n) 30 { 31 pr = st.insert(P[i].s); 32 if (pr.second) 33 { 34 strcpy(S[cnt ++].s, P[i].s); 35 continue; 36 } 37 while (i <= m) 38 { 39 ++ i; 40 pr = st.insert(P[i].s); 41 if (pr.second) 42 { 43 strcpy(S[cnt ++].s, P[i].s); 44 break; 45 } 46 } 47 } 48 49 if (cnt == 0) 50 { 51 printf("Keep going...\n"); 52 return 0; 53 } 54 for (int i = 0; i < cnt; ++ i) 55 printf("%s\n", S[i].s); 56 return 0; 57 }
转载于:https://www.cnblogs.com/GetcharZp/p/9581881.html
pat 1124 Raffle for Weibo Followers(20 分)相关推荐
- 【PAT (Advanced Level) Practice】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 ...
- PAT_A 1124. Raffle for Weibo Followers (20)
1124. Raffle for Weibo Followers (20) John got a full mark on PAT. He was so happy that he decided t ...
- PAT 1124 Raffle for Weibo Followers
PAT 1124 Raffle for Weibo Followers Java 1.题意 输入:数字m,n,s,一串人名. 第s个为获奖的人,则s+n 为下一个获奖的人,如果此人已经获奖,那么则往后 ...
- PAT 1124 Raffle for Weibo Followers python解法
1124 Raffle for Weibo Followers (20 分) John got a full mark on PAT. He was so happy that he decided ...
- PTA甲1124 Raffle for Weibo Followers (20 point(s))
强烈推荐,刷PTA的朋友都认识一下柳神–PTA解法大佬 本文由参考于柳神博客写成 柳神的CSDN博客,这个可以搜索文章 柳神的个人博客,这个没有广告,但是不能搜索 还有就是非常非常有用的 算法笔记 全 ...
- PAT-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 ...
- PAT甲级1124 Raffle for Weibo Followers :[C++题解]哈希表、微博转发抽奖
文章目录 题目分析 题目来源 题目分析 来源:acwing 分析:开一个哈希表存已经中将的用户,避免重复发奖. 遍历所有m条姓名,从第一个中奖的开始,依次模拟即可. ac代码 #include< ...
- PAT(甲)1124 Raffle for Weibo Followers——未完成
题目链接 按Ctrl单机链接,打开题目页面. 做题过程 题目分值 20 分 提交次数 分值 原因 解决方案 1 12 答案错误 重复n次 2 12 答案错误 未解决 收获:基本错题是通过样例猜题意,然 ...
- 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 ...
最新文章
- 解题报告 - 牛客练习赛63 C - 牛牛的揠苗助长(货仓选址+二分)
- 百度编辑器(ueditor)上传图片
- [书目20130216]深入浅出WPF
- Power Designer使用技巧
- 细说SSO单点登录(转)
- 自主学习之RxSwift(二) -----flatMap
- 数据流程图顶层一层二层_只需三个公式,三阶魔方超简单入门图文教程5:完全复原第二层...
- 快速对比UART、SPI、I2C通信的区别与应用
- java quartz tomcat_Quartz Scheduler - 在Tomcat或应用程序jar中运行?
- list、tuple、dict、set、map
- Android环境配置(Eclipse全开发环境下载)
- 研磨设计模式--简单工厂
- CMS-订单系统的分布式事务如何处理
- vs13配置matlab,VS配置电脑系统变量(VS2013+matlab2018a)
- Macbook变速播放视频
- 4G模块的GPS定位差距过大
- 亿级视频内容如何实时更新?
- win10 cortana搜索没有任何结果,只是一片空白的解决方案
- 用断点续存实现视频快速上传
- c语言 爱心 (koi奋斗中~~)