标签:main div from tor code bsp ecif sam mes
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! TryAgainAgainSample Output 1:
PickMe Imgonnawin! TryAgainAgainSample Input 2:
2 3 5 Imgonnawin! PickMeSample 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; } }
PAT1124:Raffle for Weibo Followers
标签:main div from tor code bsp ecif sam mes
原文地址:http://www.cnblogs.com/0kk470/p/7739808.html