toj 2485 Card Tric

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原文链接：http://www.cnblogs.com/forever4444/archive/2009/05/20/1471149.html

本文介绍了一个名为CardTric的程序，该程序用于解决一个经典的魔术问题：如何预先排列好一副扑克牌，使得按照特定规则依次翻开时，每张牌都恰好符合预期的顺序。文章提供了完整的代码实现，并附带了1到13张牌的初始排列解决方案。

2485. Card Tric

Time Limit: 0.5 Seconds Memory Limit: 65536K
Total Runs: 348 Accepted Runs: 200

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 specifications

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 specifications

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

Output for sample input

2 1 4 3
3 1 4 5 2

Source: Nordic Collegiate Contest 2006

Submit List Runs Forum Statistics

#include < iostream >
#include < string >
using namespace std;
string str[ 14 ];
int t,n;
int main()
{
    cin >> t;
    str[ 1 ] = " 1 " ;
    str[ 2 ] = " 2 1 " ;
    str[ 3 ] = " 3 1 2 " ;
    str[ 4 ] = " 2 1 4 3 " ;
    str[ 5 ] = " 3 1 4 5 2 " ;
    str[ 6 ] = " 4 1 6 3 2 5 " ;
    str[ 7 ] = " 5 1 3 4 2 6 7 " ;
    str[ 8 ] = " 3 1 7 5 2 6 8 4 " ;
    str[ 9 ] = " 7 1 8 6 2 9 4 5 3 " ;
    str[ 10 ] = " 9 1 8 5 2 4 7 6 3 10 " ;
    str[ 11 ] = " 5 1 6 4 2 10 11 7 3 8 9 " ;
    str[ 12 ] = " 7 1 4 9 2 11 10 8 3 6 5 12 " ;
    str[ 13 ] = " 4 1 13 11 2 10 6 7 3 5 12 9 8 " ;
     while (t -- )
    {
        cin >> n;
        cout << str[n] << endl;
    }
     return 0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/20/1471149.html