pat 1124 Raffle for Weibo Followers(20 分)

时间:2023-03-08 16:24:10
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...
 #include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <set>
using namespace std;
const int MAX = 1e3 + ; int m, n, s, cnt = ;
struct node
{
char s[];
}P[MAX], S[MAX];
set <string> st;
pair <set <string> :: iterator, bool> pr;
set <string> :: iterator iter; int main()
{
// freopen("Date1.txt", "r", stdin);
scanf("%d%d%d", &m, &n, &s);
for (int i = ; i <= m; ++ i)
scanf("%s", &P[i].s); for (int i = s; i <= m; i += n)
{
pr = st.insert(P[i].s);
if (pr.second)
{
strcpy(S[cnt ++].s, P[i].s);
continue;
}
while (i <= m)
{
++ i;
pr = st.insert(P[i].s);
if (pr.second)
{
strcpy(S[cnt ++].s, P[i].s);
break;
}
}
} if (cnt == )
{
printf("Keep going...\n");
return ;
}
for (int i = ; i < cnt; ++ i)
printf("%s\n", S[i].s);
return ;
}