PAT1124:Raffle for Weibo Followers

Posted 2020-10-12 0kk470

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 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

1.map模拟一个字典dic记录该用户是否领过奖。
2.用一个bool值hasWinner来标记是否有人获过奖。

代码

#include<iostream>
#include<vector>
#include<map>
using namespace std;
int main()
{
    int M,N,S;
    while(cin >> M >> N >> S)
    {
       vector<string> List(M + 1);
       map<string,int> dic;
       bool hasWinner = false;
       for(int i = 1;i <= M;i++)
       {
           cin >> List[i];
       }

       for(int i = S;i <= M;i += N)
       {
           while(dic.count(List[i]) > 0 && i <= M) i++;
           if(i > M) break;
           hasWinner = true;
           cout << List[i] << endl;
           dic[List[i]]++;
       }
       if(!hasWinner)
          cout << "Keep going..." << endl;
    }
}