卡片游戏 （Throwing cards away I） UVa 10935

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.

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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

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题目不长，应该很好理解，这是一道运用到队列的题目，废话不多说，下面贴出代码：

这里需要注意的是当 n为1 的时候应当输出的是：

Discarded cards: 

Remaining card: 1

#include
#include
using namespace std;
int main()
{
	int n;
	while(cin >> n && n){
		queue s;
		for(int i=1; i<=n; ++i)
			s.push(i);
		
		if( n == 1 )	cout << "Discarded cards:" << endl;
		else	cout << "Discarded cards:" << " " << s.front();
		if( n == 2 )	cout << endl;
		while( n-1 ){
			s.pop();  n--;
			int x = s.front();
			s.pop();
			s.push(x);
			if( n >= 2  )cout << "," << " " << s.front();
			if( n == 2 )	cout << endl;	
		}
		cout << "Remaining card:" << " " << s.front() << endl;	
	}
	return 0;
}

所以只需要讨论一下，控制一下条件即可。