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<bits/stdc++.h> using namespace std; typedef long long ll; int main(){ int m,n,s; cin >> m >> n >> s; vector<string> vec(m+1); for(int i=1;i <= m;i++){ string t;cin >> t; vec[i] = t; } if(s > m) {printf("Keep going...\n");return 0;} map<string,int> mp; for(int i=s;i <= m;){ if(mp[vec[i]] == 0){ cout << vec[i] << endl; mp[vec[i]] = 1; i += n; } else{ i++; } } return 0; }
发现一个性质,题面越长一般越简单。。