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 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...

Keys:

• 模拟题

Code:

 /*
 Data: 2019-07-04 15:00:05
 Problem: PAT_A1124#Raffle for Weibo Followers
 AC: 25:52
 
 题目大意：
 从转发微博的名单中抽奖，每隔N人送一份奖品
 输入：
 第一行给出，转发数M<=1000，获奖者间隔的人数N，第一个获奖者序号S>=1
 接下来M行，转发人的昵称
 注：若选定的获奖者如果已经获奖，则考虑下一个人
 输出：
 打印获奖者名单
 
 基本思路：
 从S开始遍历名单，计数器统计跳过人数，跳过n人时进行查询
 若未获奖，计数器清零，打印姓名
 若已获奖，计数器不变，查询下一位
 */
 #include<cstdio>
 #include<string>
 #include<map>
 #include<iostream>
 using namespace std;
 const int M=1e3+;
 string info[M];
 map<string,int> mp;
 
 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif
 
     int m,n,s,skip=;
     scanf("%d%d%d", &m,&n,&s);
     for(int i=; i<=m; i++)
         cin >> info[i];
     for(int i=s; i<=m; i++)
     {
         if(i==s || skip==n)
         {
             if(mp[info[i]]==){
                 cout << info[i] << endl;
                 mp[info[i]]=;
                 skip=;
             }
             else
                 continue;
         }
         skip++;
     }
     if(s > m)
         printf("Keep going...");
 
     return ;
 }