【PAT】【Advanced Level】1124. Raffle for Weibo Followers (20)

1124. Raffle for Weibo Followers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 2Imgonnawin!PickMePickMeMeMeeeLookHereImgonnawin!TryAgainAgainTryAgainAgainImgonnawin!TryAgainAgain

Sample Output 1:

PickMeImgonnawin!TryAgainAgain

Sample Input 2:

2 3 5Imgonnawin!PickMe

Sample Output 2:

Keep going...

原题链接：

https://www.patest.cn/contests/pat-a-practise/1124

思路：

map映射

遇到领过的加1，跳过。否则领奖，加N

CODE：

#include<iostream>#include<cstring>#include<string>#include<map>#include<vector>using namespace std;vector<string> vec;map<string,int> ma;int main(){int n,m,s;cin>>n>>m>>s;for (int i=0;i<n;i++){string t;cin>>t;vec.push_back(t);}int st=s-1;int num=0;while (st<n){if (ma[vec[st]]==0){ma[vec[st]]=1;cout<<vec[st]<<endl;num++;st+=m;}else{st++;}}if (num==0){cout<<"Keep going..."<<endl;}return 0;}