PAT 1124 Raffle for Weibo Followers python解法

1124 Raffle for Weibo Followers （20 分）
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…

题意：模拟微博抽奖，第一行给出3个数（m,n,s），分别表示参与抽奖人数，获奖人数，抽奖规则（隔s个人抽一个），如果原来被抽取过，则顺延到下一个。

解题思路：用列表candidate来保存获奖者昵称，首先判断参与抽奖人数n是否小于抽奖规则s，若小于则没有获奖者，输出Keep going…，否则先将s-=1（因为列表索引从0开始），开始循环判断昵称是否在candidate中，如果在，则s+=1，不在，则判断是否满足抽奖规则（i==s），满足则将昵称添加到candidate中，并将s+=n。

m, n, s = map(int,input().split())
#m, n, s = 9,3,2
#m, n, s = 2,3,5
l = []
for i in range(m):l.append(input())
#l = ['Imgonnawin!','PickMe']
#l = ['Imgonnawin!','PickMe','PickMeMeMeee','LookHere','Imgonnawin!','TryAgainAgain','TryAgainAgain','Imgonnawin!','TryAgainAgain']
candidate = []
if s>n:print('Keep going...')
else:s -= 1for i in range(m):if l[i] in candidate:s += 1 if i == s and l[i] not in candidate:candidate.append(l[i])s += n       print('\n'.join(candidate))

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