1124 Raffle for Weibo Followers[简单]

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.

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...

题目大意：输出m,n，s分别表示转发微博的人数，寻找的跳数，初始winner，并且要求如果已经获过奖的就不能再获奖，下次遍历到它时就应该跳过。

//第一次我是这么想的，对所有重复转发的只记录第一次的转发，但是后来发现，这样:

因为初始获奖者跳过了Imgonnawin!用户，下一次再出现时我没有将其记录，就顺次到了下一位用户。这就出现了错误。

AC代码:

#include <iostream>
#include <map>
#include <cstdio>
#include <vector>
using namespace std;
map<string,int> mp;
vector<string> vt;
int main() {
    int m,n,s;
    cin>>m>>n>>s;
    string name;
    for(int i=;i<=m;i++){
        cin>>name;
        vt.push_back(name);//存都是要存起来的。
    }
   // cout<<endl;
    int len=vt.size();
    bool flag=false;
    for(int i=s-;i<len;i+=n){
        while(mp.count(vt[i])!=){
            i++;
        }
        if(mp.count(vt[i])==){//如果它没有出现过
            flag=true;
            cout<<vt[i]<<'\n';
            mp[vt[i]]=;//出现过。
        }
    }

    if(!flag)cout<<"Keep going...";
    return ;
}

//这个题目真的是比较简单的，如果当前已经领过奖了，那么就一直跳过，如果没有出现过就输出并且标记，算是一星的难度吧。