河南省第六届大学生程序设计竞赛 F Card Trick

1627: Card Trick

时间限制: 1 Sec  内存限制: 128 MB
提交: 37  解决: 25
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题目描述

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

245

样例输出

2 1 4 33 1 4 5 2

提示

来源

河南省第六届大学生程序设计竞赛

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题意：给你n张牌，让你变一个魔术：第1次把上面的1张牌放到底部，然后最上面的牌就是1，然后拿走1。第2次把上面的2张牌依次放到底部，然后最上面的牌就是2，然后拿走2....重复这个过程，直到所有的牌都被拿走。问一开始的牌应该从上到下怎么放，才能完成这个魔术。

算法：逆向思维，从后向前模拟。

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#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

  

using namespace std;

  

int n;

queue<int> q;

int f;

  

void output()  /// 递归输出q里面的内容

{

  

    inttmp=q.front();

    q.pop();

    while(!q.empty())

         output();

  

    if(f==0)

    {

       printf("%d",tmp);

       f++;

    }

    else

        printf(" %d",tmp);

}

  

int main()

{

    intt;

    scanf("%d",&t);

    while(t--)

    {

        f=0;

        scanf("%d",&n);

        intcnt;

        while(n>0)

        {

            q.push(n);

            cnt=n;

            n--;

            while(cnt--) /// 由后向前翻cnt次牌

            {

                inttmp=q.front();

                q.pop();

                q.push(tmp);

            }

        }

        output();

        printf("\n");

    }

    return0;

}