PAT甲级 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...

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题目的意思是给出n个人名字，每个m个人抽奖和抽奖起始点st，输出中奖的人名字，如果重复则下一个

思路：直接暴力，将中奖的人名字扔到一个set里面，每次先判断

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<string>#include<queue>#include<stack>#include<map>#include<set>using namespace std;#define LL long longconst int inf=0x3f3f3f3f;string str[1005];int main(){    int n,m,st;    scanf("%d%d%d",&n,&m,&st);    for(int i=1; i<=n; i++)        cin>>str[i];    set<string>s;    s.clear();    int flag=0;    for(int i=st; i<=n; i+=m)    {        while(s.count(str[i])==1&&i<=n)        {            i++;        }        if(i>n)            break;            s.insert(str[i]);            flag=1;            cout<<str[i]<<endl;    }    if(flag==0)        cout<<"Keep going..."<<endl;    return 0;}