PAT-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 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...

---

提交代码

对这道题的题意有些误解，我以为是在抽到奖时看这个人是否得过奖，而实际上是每次叠加人数时，就要判断这个人是否得过奖，得了奖就忽略。

#include <bits/stdc++.h>

using namespace std;

int N, M, K;
vector<string> userVec;
vector<string> ansVec;

int main()
{
    cin>>N>>M>>K;
    string user;
    int m = 0;
    int k = 1;
    for(int i = 1; i <= N; i++) {
        cin>> user;
        //if(i >= K+m*M) {
            int flag = 1;
            for(int j = 0; j < ansVec.size(); j++) {
                if(user == ansVec[j]) {
                    flag = 0;
                    break;
                }
            }
            if(flag) {
                if(k >= K+m*M) {
                    ansVec.push_back(user);
                    m++;
                }
                k++;
            }
        //}
    }
    if(ansVec.empty()) {
        cout<< "Keep going..."<< endl;
        return 0;
    }
    for(int i = 0; i < ansVec.size(); i++) {
        cout<< ansVec[i]<< endl;
    }
    return 0;
}