PAT甲级 1124. Raffle for Weibo Followers (20)

微博抽奖程序设计

最新推荐文章于 2022-09-03 19:12:35 发布

原创最新推荐文章于 2022-09-03 19:12:35 发布 · 439 阅读

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本文介绍了一个简单的抽奖程序设计案例，该程序用于从微博转发者中按照特定规则抽取幸运粉丝，并确保每位用户仅能中奖一次。

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 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 long
const 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;
}