POJ,3032,Card Trick

最新推荐文章于 2019-04-04 10:46:36 发布

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最新推荐文章于 2019-04-04 10:46:36 发布

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分类专栏：技巧模拟 OJ 文章标签： ACM 编程 POJ 队列

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本文介绍了一个名为CardTrick的编程题目，该题目要求通过特定的算法来确定初始的卡片排列顺序。文章提供了一份C语言实现代码，并分享了解题过程中遇到的一些挑战，如数组大小的调整等。

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Card Trick

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3530		Accepted: 2525

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

Source

Nordic 2006

#include <stdio.h>
#include <string.h>
int main()
{
	int *q[200];
	int a[200],n,x,i,j,t,h;
	scanf("%d",&n);
	while (n--)
	{
		scanf("%d",&x);
		memset(a,0,sizeof(a));
		for (i=1;i<=x;i++)
			q[i]=&a[i];
		h=1;
		t=x;
		for (i=1;i<=x;i++)
		{
			for (j=1;j<=i;j++)
			{
				q[++t]=q[h++];
			}
			*q[h++]=i;
		}
		for (i=1;i<x;i++)
			printf("%d ",a[i]);
		printf("%d\n",a[i]);
	}
	return 0;
}

气死我了，我说我这么明确的思路怎么会出错呢？谁知道数组开小了，13个数，开到100居然不够，居然让我整了一晚上

就是建一个存整型指针的数组，与整形数组一一对应后，让数组所存的指针进出队，最后将队头指针所指向的地址的内容赋值