NYOJ_714_Card Trick【隊列】

/*Card Trick
時間限制：1000 ms  |  內存限制：65535 KB
難度：3

描述

    The magician shuffles a small pack of cards, holds it face down and performs the following procedure:

        The top card is moved to the bottom of the pack. The new top card is dealt face up onto the table. It is the Ace of Spades.
        Two cards are moved one at a time from the top to the bottom. The next card is dealt face up onto the table. It is the Two of Spades.
        Three cards are moved one at a time…
        This goes on until the nth and last card turns out to be the n of Spades.

    This impressive trick works if the magician knows how to arrange the cards beforehand (and knows how to give a false shuffle). Your program has to determine the initial order of the cards for a given number of cards, 1 ≤ n ≤ 13.

輸入
    On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 10 Each case consists of one line containing the integer n. 1 ≤ n ≤ 13
輸出
    For each test case, output a line with the correct permutation of the values 1 to n, space separated. The first number showing the top card of the pack, etc…
樣例輸入

    2
    4
    5

樣例輸出

    2 1 4 3
    3 1 4 5 2

*/   

題意：  抽取上面第一張牌後放到最後，然後現在上面顯現的是（梅花）A，然後去掉A，然後再抽取兩張放到後面，現在上面的牌顯示（梅花） 2 ， 然後去掉2  ，依次，，，知道第n次 。。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
queue<int> Q;
int s[30];
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int n;
		int i,j,k=1;
		int index = 0;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		{
			Q.push(i);
		}
		for(i=1;i<=n;i++)
		{
			j = i;
			while(j--)//隊列 彈出次數 1.2.3...n
			{
				int t = Q.front();
				Q.pop();
				
				Q.push(t);
			}
			index = Q.front();//隊首 彈出 index是指 哪個位置 1.2.3...n
			Q.pop();
			s[index] = k++;
		}
		for(i=1;i<=n;i++)
			printf("%d ",s[i]);
		printf("\n");
	}
	return 0;
}