Uva

Problem B: 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 numbern ≤ 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

用queue直接解决，熟练运用队列操作

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>

using namespace std;

int main()
{
	int n;
	while (cin >> n && n) {
		queue<int> cards;
		// 初始化队列
		for (int i = 1; i <= n; i++) {
			cards.push(i);
		}
		printf("Discarded cards:");
		int first = 1;

		while (cards.size() != 1) {
			if (first) {
				printf(" %d", cards.front());
				first = 0;
			}
			else {
				printf(", %d", cards.front());
			}
			cards.pop();
			// 新的第一张牌先出队列再入队列，记得先存贮
			int t = cards.front();
			cards.pop();
			cards.push(t);
		}
		printf("
Remaining card: %d
", cards.front());
	}

	return 0;
}