PAT (Advanced Level) Practise 1124 Raffle for Weibo Followers (20)

1124. Raffle for Weibo Followers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 2Imgonnawin!PickMePickMeMeMeeeLookHereImgonnawin!TryAgainAgainTryAgainAgainImgonnawin!TryAgainAgain

Sample Output 1:

PickMeImgonnawin!TryAgainAgain

Sample Input 2:

2 3 5Imgonnawin!PickMe

Sample Output 2:

Keep going...

题意：抽奖活动，一共m个人，按顺序每隔n个人就中奖。可能有人转发多次，但不能中奖多次。如果处于当前中奖位置的网友已经中过奖，则跳过他顺次取下一位。确定中奖名单。如果没有人中奖，则输出“Keep going...”

解题思路：模拟

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m, x;string s[1005];map<string, int>mp;int main(){while (~scanf("%d%d%d", &n, &m, &x)){for (int i = 1; i <= n; i++) cin >> s[i];mp.clear();if (x > n) puts("Keep going...");else{for (int i = x; i <= n;){if (mp[s[i]]) { i++; continue; }mp[s[i]] = 1; cout << s[i] << endl; i += m;}}}return 0;}