1124. Raffle for Weibo Followers (20)

最新推荐文章于 2022-11-09 21:26:16 发布

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本文介绍了一个抽奖程序的设计与实现，该程序能够从微博转发者中按照指定的间隔选择幸运儿，并确保每位用户仅能获奖一次。文章提供了完整的代码实现，演示了如何通过输入转发总数、间隔数及首位幸运儿的位置来生成获奖名单。

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...

代码实现

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
int main()
{
  int n,m,s;
  bool flag=false;
  map<string,int>mapp;
  cin>>m>>n>>s;
  char people[1005][25];
  for(int i=1;i<=m;i++)
  {
      scanf("%s",people[i]);
  }
  for(int i=s;i<=m;i=i+n)
  {
      if(mapp.find(people[i])==mapp.end())
      {
          mapp[people[i]]=1;
          printf("%s\n",people[i]);
          flag=true;
          continue;
      }
      if(mapp[people[i]]==1)
      {
          while(mapp[people[i]]==1 && i<=m)
          {
              i++;
          }
          mapp[people[i]]=1;
          printf("%s\n",people[i]);
          flag=true;
      }
  }
  if(flag==false)
  {
      printf("Keep going...\n");
  }
  return 0;
}