Source:
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:
1 /*
2 Data: 2019-07-04 15:00:05
3 Problem: PAT_A1124#Raffle for Weibo Followers
4 AC: 25:52
5
6 题目大意：
7 从转发微博的名单中抽奖，每隔N人送一份奖品
8 输入：
9 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
10 接下来M行，转发人的昵称
11 注：若选定的获奖者如果已经获奖，则考虑下一个人
12 输出：
13 打印获奖者名单
14
15 基本思路：
16 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
17 若未获奖，计数器清零，打印姓名
18 若已获奖，计数器不变，查询下一位
19 */
20 #include<cstdio>
21 #include<string>
22 #include<map>
23 #include<iostream>
24 using namespace std;
25 const int M=1e3+10;
26 string info[M];
27 map<string,int> mp;
28
29 int main()
30 {
31 #ifdef ONLINE_JUDGE
32 #else
33 freopen("Test.txt", "r", stdin);
34 #endif
35
36 int m,n,s,skip=0;
37 scanf("%d%d%d", &m,&n,&s);
38 for(int i=1; i<=m; i++)
39 cin >> info[i];
40 for(int i=s; i<=m; i++)
41 {
42 if(i==s || skip==n)
43 {
44 if(mp[info[i]]==0){
45 cout << info[i] << endl;
46 mp[info[i]]=1;
47 skip=0;
48 }
49 else
50 continue;
51 }
52 skip++;
53 }
54 if(s > m)
55 printf("Keep going...");
56
57 return 0;
58 }
原文地址：https://www.cnblogs.com/blue-lin/p/11132783.html