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  • 1124 Raffle for Weibo Followers (20 分)

    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...
    
     
    简单标记即可
     
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 int n,m,k;
     4 string s[1005];
     5 map<string,int> mp;
     6 int main(){
     7     cin >> n >> m >> k;
     8     for(int i = 1 ; i <= n; i++){
     9         cin >> s[i];
    10     }
    11     if(k > n){
    12         cout<<"Keep going..."<<endl;
    13     }else{
    14         while(k <= n){
    15             if(mp[s[k]] == 0){
    16                 mp[s[k]] = 1;
    17                 cout << s[k] << endl;
    18                 k += m;
    19             }else{
    20                 k++;
    21             }
    22         }
    23     }
    24     return 0;
    25 }
     
     
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/zllwxm123/p/11320347.html
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