1124 Raffle for Weibo Followers (20 分)
标签：code The fir lan happy app select div lex
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.
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.
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.
9 3 2
Imgonnawin!
PickMe
PickMeMeMeee
LookHere
Imgonnawin!
TryAgainAgain
TryAgainAgain
Imgonnawin!
TryAgainAgain
PickMe
Imgonnawin!
TryAgainAgain
2 3 5
Imgonnawin!
PickMe
Keep going...
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,m,k; 4 string s[1005]; 5 map<string,int> mp; 6 int main(){ 7 cin >> n >> m >> k; 8 for(int i = 1 ; i <= n; i++){ 9 cin >> s[i]; 10 } 11 if(k > n){ 12 cout<<"Keep going..."<<endl; 13 }else{ 14 while(k <= n){ 15 if(mp[s[k]] == 0){ 16 mp[s[k]] = 1; 17 cout << s[k] << endl; 18 k += m; 19 }else{ 20 k++; 21 } 22 } 23 } 24 return 0; 25 }
1124 Raffle for Weibo Followers (20 分)
标签：code The fir lan happy app select div lex
原文地址：https://www.cnblogs.com/zllwxm123/p/11320347.html
