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.
微博抽奖，从第一个中奖的编号开始，每隔N个人中奖，如果第 (S+N) 个人已经中过奖，则检查编号 (S+N+1)。如果S+N+1没有中过奖，则从S+N+1开始，再隔N个人中奖
如果理解为编号S没有中奖，考查S+1后，不改变S，还是从S+N抽奖，那么测试点3就错误
//略微修改柳神代码 #include <iostream> #include <map> using namespace std; int main() { int M, N, S; scanf("%d%d%d", &M, &N, &S); string str; map<string, int> str_to_int; bool flag = false; for (int i = 1; i <= M; i++) { cin >> str; // if (str_to_int[str] == 1) s = s + 1; if (i == S && str_to_int[str] == 0) { //到中奖编号，且没有中过奖 str_to_int[str] = 1; cout << str << endl; flag = true; S = S + N; } else if (i == S && str_to_int[str] == 1) {//已经中过奖，考查S+1 S = S + 1; } } if (!flag) cout << "Keep going..."; return 0; }