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

题意：给定你一个数n，牌从1~n按顺序排列，首先把第一张牌扔掉。然后再把最上面的放到最后面。

依次循环，直到仅仅有一张牌为止。第一行输出按顺序扔掉的牌，第二行输出最后剩下的牌。

思路：本题须要用到数据结构---队列，先让第一个为head，最后一个元素的后一个为tail把第一个元素向前移动一个位置，这样就删除了第一个，然后再把第一个元素移到最后面。tail向后移动一个位置。

代码：

#include<cstdio>

using namespace std;

int q[200];

int main()

{

    int head,tail,x;

    while(scanf("%d",&x)!=EOF&&x!=0)

    {

        for(int i=1;i<=x+1;i++)

        q[i]=i;

        head=1;

        tail=x+1;

        printf("Discarded cards:");

        while(head<tail-1)

        {

        if(head<tail-2)

          printf(" %d,",q[head]);

          else

            printf(" %d",q[head]);

             head++;

             q[tail]=q[head];

        tail++;

          head++;

        }

        printf("\n");

       printf("Remaining card:");

        printf(" %d",q[head]);

        printf("\n");

    }

    return 0;

}