POJ 3032 Card Trick（较好模拟题，双向队列deque）

最新推荐文章于 2019-01-08 01:53:01 发布

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最新推荐文章于 2019-01-08 01:53:01 发布

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分类专栏：模拟题

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本文介绍了一个名为CardTrick的编程挑战问题，该问题要求根据特定的发牌规则逆向推导出初始的牌堆排列。通过分析题目要求及给出的示例输入输出，提供了一段使用C++编写的解决方案代码，利用deque容器实现了解题算法。

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1、http://poj.org/problem?id=3032

2、

Card Trick

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3231		Accepted: 2359

Description

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.

Input

On the first line of the input is a single positive integer, telling the number of test cases to follow. Each case consists of one line containing the integer n.

Output

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…

Sample Input

2
4
5

Sample Output

2 1 4 3
3 1 4 5 2

3、代码：

#include<stdio.h>
#include<deque>
#include<algorithm>
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        deque<int> q;
        q.push_back(n);
        for(int i=n-1;i>=1;i--)
        {
            q.push_back(i);
            for(int j=1;j<=i;j++)
            {
                int temp=q.front();
                q.push_back(temp);
                q.pop_front();
            }
        }
        while(!q.empty())
        {
            printf("%d ",q.back());
            q.pop_back();
        }
        printf("\n");
    }
    return 0;
}