Source:

PAT A1124 Raffle for Weibo Followers (20 分)

Description:

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...

Keys:

  • 模拟题

Code:

 /*
Data: 2019-07-04 15:00:05
Problem: PAT_A1124#Raffle for Weibo Followers
AC: 25:52 题目大意:
从转发微博的名单中抽奖,每隔N人送一份奖品
输入:
第一行给出,转发数M<=1000,获奖者间隔的人数N,第一个获奖者序号S>=1
接下来M行,转发人的昵称
注:若选定的获奖者如果已经获奖,则考虑下一个人
输出:
打印获奖者名单 基本思路:
从S开始遍历名单,计数器统计跳过人数,跳过n人时进行查询
若未获奖,计数器清零,打印姓名
若已获奖,计数器不变,查询下一位
*/
#include<cstdio>
#include<string>
#include<map>
#include<iostream>
using namespace std;
const int M=1e3+;
string info[M];
map<string,int> mp; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif int m,n,s,skip=;
scanf("%d%d%d", &m,&n,&s);
for(int i=; i<=m; i++)
cin >> info[i];
for(int i=s; i<=m; i++)
{
if(i==s || skip==n)
{
if(mp[info[i]]==){
cout << info[i] << endl;
mp[info[i]]=;
skip=;
}
else
continue;
}
skip++;
}
if(s > m)
printf("Keep going..."); return ;
}

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