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...
题目大意:在转发转发微博的人(一堆名字)中按照设定好的规则选择中奖者;M是转发的人数,N是选择winner时间隔的人数,S是第一个winner的index。。。
思路:把名字依次放入字符串数组name中,遍历name,若当前的转发者不在集合se中,则为winner,输出人名并且将它放入se中;若当前的转发者在se中,则往后找name;遍历name之后如果se为空,则输出“Keep going...”
#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
using namespace std;
int main()
{
int M, N, S;
scanf("%d%d%d", &M, &N, &S);
unordered_set<string> se;
vector<string> name;
for (int i = 0; i < M; i++) {
string tmp;
cin >> tmp;
name.push_back(tmp);
}
for (int i = S-1; i < name.size(); i = i + N) {
while (1) {
if (i >= name.size())
break;
if (se.find(name[i]) == se.end())
break;
i++;
}
if (i < name.size()) {
cout << name[i] << endl;
se.insert(name[i]);
}
}
if (se.empty())
printf("Keep going...\n");
}