E Card Trick(队列)

                                                                                  Card Trick

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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

1. 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.
2. 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.
3. Three cards are moved one at a time…
4. 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.

输入

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

输出

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…

样例输入

2 4 5

样例输出

2 1 4 3 

3 1 4 5 2

大致题意：

一沓牌，把最上面的一张放在最下面，此时最上面的牌是黑桃A
接着把上面的二张牌放在最下面，此时最上面的牌是黑桃2
接着把上面的三牌放在最下面，此时最上面的牌是黑桃3
........
接着把上面的n张牌放在最下面，此时最上面的牌是黑桃n
求原始序列

解题思路：

用队列模拟过程，首先存入队列的为数组下标，然后进行一次操作，取队首元素，此下标数组存的为操作的编号

我的代码：

#include<bits/stdc++.h>using namespace std;int main(){    int num;    scanf("%d",&num);    while(num--)    {        int a[15]= {0},n,i,j,t;        scanf("%d",&n);        queue<int>q;        for(i=1; i<=n; i++)//数组下标进队列            q.push(i);        int ok=1;        for(i=1; i<=n; i++)        {            j=i;            while(j--)//把j个牌放在底部            {                t=q.front();                q.pop();                q.push(t);            }            a[q.front()]=ok;//抽取最上面的一个为上面显示的是ok            q.pop();            ok++;        }        for(i=1;i<n;i++)            printf("%d ",a[i]);        printf("%d\n",a[n]);    }    return 0;}