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PAT--1124. Raffle for Weibo Followers

2019-11-06 06:57:06
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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…

题解

#include <iostream>#include <string>#include <algorithm>#include <vector>#include <cstring>#include <cstdio>#include <set>using namespace std;const int maxn = 1000 + 5;int m, n, s;string candidates[maxn];int main(){#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);#endif // ONLINE_JUDGE cin >> m >> n >> s; for(int i = 1; i <= m; ++i){ cin >> candidates[i]; } vector<string> ans; set<string> Set; for(int i = s; i <= m;){ if(Set.count(candidates[i]) == 0){ ans.push_back(candidates[i]); Set.insert(candidates[i]); i += n; }else{ i++; } } if(ans.empty()) cout << "Keep going..." << endl; else{ for(string s : ans) cout << s << endl; } return 0;}
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