pat 1124 Raffle for Weibo Followers（20 分）

Posted 2021-01-03 getcharzp

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...
");
52         return 0;
53     }
54     for (int i = 0; i < cnt; ++ i)
55         printf("%s
", S[i].s);
56     return 0;
57 }