Uva 10935 Throwing cards away I

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 find the sequence of discarded cards and the last, remaining card.

Each line of input (except the last) contains a number n ≤ 50. The last line contains 0 and this line should not be processed. For each number from the input produce two lines of output. The first line presents the sequence of discarded cards, the second line reports 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

Output for sample input

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

这道题就是一个队列的问题，之所以把这道题拉上来，是想说这个题PE的事。 

我之前把回车换行放在了discarded cards的数字后面，结果一直PE。因为这样n=1的时候，输出会变成一排不会输出一个回车换行。不过这道题居然也没有提n=1的特殊情况应该怎么输出，只能去尝试它到底是什么格式了.....

#include 
#include 
#include 
#include 
#include 
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        queue que;
        int a[55];
        for(int i=1;i<=n;i++)
            que.push(i);
        for(int i=1;i<=n-1;i++)
        {
            int x=que.front();
            que.pop();
            a[i]=x;
            x=que.front();
            que.pop();
            que.push(x);
        }
        int y=que.front();
        printf("Discarded cards:");
        for(int i=1;i<=n-1;i++)
        {
            if(i!=n-1) printf(" %d,",a[i]);
            else printf(" %d",a[i]);
        }
        printf("\nRemaining card: %d\n",y);
    }
    return 0;
}