UVA-10935

Description

 

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

---

Folklore, adapted by Piotr Rudnicki

题目大意：从1到n的有序卡片，执行如下操作：

从最上层取一张一处牌堆，再从最上层取一张放置到最底层

直到牌堆只剩一张牌为止。

输出移除卡牌的顺序和最后保留的卡片

解题思路：队列

WA点：当n=1时没有移除卡牌

代码：

#include 
#include 
#include 

using namespace std;

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