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...

#include<stdio.h>#include<iostream>#include<string>#include<map>using namespace std;map<string,int>cou;int main(){int m,n,s,i,flag=0;scanf("%d %d %d",&m,&n,&s);string name[1010];for(i=1;i<=m;i++){cin>>name[i];}for(i=s;i<=m;i=i+n){if(cou[name[i]]==0){cou[name[i]]=1;cout<<name[i]<<endl;flag=1;}else{i=i-n+1;}}if(flag==0){printf("Keep going...");}}