UVA-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.

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 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

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

题意：有n张的纸牌，初始的时候是按照顺利排列的，从上至下依次是1、2、3….n，接下来执行以下的操作直到只剩下一张纸牌

1. 移除最上方的纸牌
2. 将下一张纸牌移动到最后
3. 继续操作1，直到剩下一张纸牌

方法：使用队列来处理这种操作，将纸牌按照顺序依次插入到队列中，依次取出队列中的元素，执行上述操作

#include 
#include 
#include 

const int MAXLEN = 55;

using namespace std;

int main()
{
    int n;
    while(scanf("%d", &n) == 1 && n)
    {
        queue<int> Cards;
        for(int i=1; i<=n; i++)
            Cards.push(i);
        printf("Discarded cards:");
        int temp;
        while(1)
        {
            if(Cards.size() == 1)
                break;
            printf(" %d", Cards.front());
            Cards.pop();
            temp = Cards.front();
            Cards.pop();
            Cards.push(temp);
            if(Cards.size() == 1)
                break;
            printf(",");
        }
        printf("\nRemaining card: %d\n", Cards.front());
    }
    return 0;
}