Throwing cards away I （队列）

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.

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

source

UVA - 10935 

题意是这样的：例如  给个7，是这样的队列

1 2 3 4 5 6 7  ，然后1输入并出队，2出队去队尾变成 3 4 5 6 7 2 以此类推 一直到队列中只剩下一个人 时结束并输出

#include
#include
using namespace std;
int n;
int main()
{

while(cin>>n,n)
{
queueq;
if(n==1)
{
cout<<"Discarded cards:"<cout<<"Remaining card: 1"<continue;
}
for(int i=2;i<=n;i++)
{
q.push(i);
}
cout<<"Discarded cards: 1";
while(q.size()>1)
{
q.push(q.front());
q.pop();
cout<<","<<" "<q.pop();

}
cout<cout<<"Remaining card: "<}
}