1124. Raffle for Weibo Followers (20)

時間限制

400 ms

記憶體限制

65536 kB

程式碼長度限制

16000 B

判題程式

Standard

作者

CHEN, Yue

John got a full mark on PAT. He was so happy that he decided to hold a raffle（抽獎） 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...

思路：定義一個vector存放被選中的人，將所有follower儲存到一個string陣列，然後遍歷這個陣列，如果vector中存在這個人，就跳到下一個人。選中一個人後，改變s為s+n。

#include<bits/stdc++.h> using namespace std; vector<string> award; //map<string,int> mapp; string fls[1006]; int m,n,s; int _find(string aim){ for(int i = 0; i < award.size(); i++){ if(aim == award[i]) return 1; } return 0; } int main(){ cin>>m>>n>>s; for(int i = 1; i <= m; i++) cin>>fls[i]; while(s <= m){ //int i = 0; while(s<= m && _find(fls[s]))s++; if(s <= m) award.push_back(fls[s]); s += n; } if(award.size() == 0){ cout<<"Keep going..."<<endl; return 0; } for(int i = 0; i < award.size(); i++) cout<<award[i]<<endl; //cout<<mapp.size()<<endl; } /* 5 1 1 sss sss sss sss sss sss */ 

最後說一下，map竟然可以直接map[key] = value這樣用，看了別人的程式碼漲姿勢了~