PAT (Advanced Level) 1124. Raffle for Weibo Followers (20) 解题报告

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...

代码：

#include <cstdio>#include <iostream>#include <algorithm>#include <map>#include <queue>#include <string>#include <vector>using namespace std;int main(){    char name[1010][50];    int m, n, s, f = 0;    map<string, int> M;    scanf("%d%d%d", &m, &n, &s);    for(int i = 1; i <= m; i++)        scanf("%s", name[i]);    for(int i = s; i <= m; i += n)    {        if(!M[name[i]])        {            M[name[i]] = 1;            printf("%s\n", name[i]);            f = 1;        }        else        {            while(i < m)            {                i++;                if(!M[name[i]])                {                    M[name[i]] = 1;                    printf("%s\n", name[i]);                    f = 1;                    break;                }            }        }    }    if(!f) printf("Keep going...\n");    return 0;}