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 }
原文地址：https://www.cnblogs.com/GetcharZp/p/9581881.html