關於佇列運用的一道簡單題

PE 兩次 注意·格式

#include <bits/stdc++.h>

using namespace std;

int main()
{
    while(1){
        int n;
        cin >> n;
        if(n==0)break;
        queue<int> q;
        for(int i=1;i<=n;i++){
            q.push(i);
        }
        cout <<"Discarded cards:";
        if(!q.empty()&&q.back()!=q.front()){
            for(int i=1;i<n;i++){
                cout <<" " << q.front();
                if(i!=n-1)cout << ",";
                q.pop();
                int temp = q.front();
                q.push(temp);
                q.pop();
            }
        //cout << endl;  如果在這個地方換行  輸入資料為 1  的時候格式錯誤
        }
        cout <<endl; // 在此處換行才AC
        cout << "Remaining card: " << q.front() << endl;
    }


    return 0;

}

原題如下：

Given is an ordered deck of n cards numbered 1
to n with card 1 at the top and card n at the
bottom. The following operation is performed as
long as there are at least two cards in the deck:
Throw away the top card and move
the card that is now on the top of the
deck to the bottom of the deck.
Your task is to nd the sequence of discarded
cards and the last, remaining card.
Input
Each line of input (except the last) contains a
number n 50. The last line contains `0' and
this line should not be processed.
Output
For each number from the input produce two
lines of output. The rst line presents the se-
quence of discarded cards, the second line re-
ports the last remaining card. No line will have
leading or trailing spaces. See the sample for the
expected format.
Sample Input
7
19
10
6
0
Sample Output
Discarded cards: 1, 3, 5, 7, 4, 2
Remaining card: 6
Discarded cards: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 4, 8, 12, 16, 2, 10, 18, 14
Remaining card: 6
Discarded cards: 1, 3, 5, 7, 9, 2, 6, 10, 8
Remaining card: 4
Discarded cards: 1, 3, 5, 2, 6
Remaining card: 4