poj 3032 Card Trick

最新推荐文章于 2019-07-17 16:59:00 发布

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最新推荐文章于 2019-07-17 16:59:00 发布

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/22/2258748.html

本文揭示了魔术师如何通过巧妙的洗牌手法，使普通的扑克牌表演出令人惊叹的特技，从Ace of Spades到King of Spades，一步步揭秘背后的数学逻辑和策略。

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Card Trick

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2838		Accepted: 2092

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

#include<iostream>
using namespace std;
int main()
{
int n,m,k;
int step;
int pointer;
int l;
int i,j;
int card[14];
    scanf("%d",&n);
for(i=0;i<n;i++)
    {
        scanf("%d",&m);
        k=1;
        step=1;
        pointer=1;
for(j=1;j<=m;j++)
            card[j]=0;
while (k<m)
        {
while(card[pointer]!=0)
                pointer=(pointer+1>m?1:pointer+1);
            l=0;
while(l!=step)
            {
if(card[pointer]==0) 
                    l++;
                pointer=(pointer+1>m?1:pointer+1);
            }
            step++;
while(card[pointer]!=0)
            {
                pointer=(pointer+1>m?1:pointer+1);
            }
            card[pointer]=k++;
        }
for(j=1;j<=m;j++)
        {
if(card[j]==0) 
            {
                card[j]=m;
break;
            }
        }    
for(j=1;j<=m-1;j++)
            printf("%d ",card[j]);
        printf("%d\n",card[m]);
    }
return 0;
}