POJ 3032 简单队列模拟

最新推荐文章于 2019-01-08 01:53:01 发布

最新推荐文章于 2019-01-08 01:53:01 发布 · 59 阅读

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#runtime

本文提供了一种解决POJ3032 CardTrick问题的方法，该问题是关于确定一副牌的初始排列，使得按照特定规则发牌时能依次得到从A到N的梅花牌。通过使用队列和自定义排序，程序能够为1至13张不同数量的牌找到正确的排列。

http://poj.org/problem?id=3032

提交的时候正赶上系统故障，全是Runtime Error!!!

一看Online Status,原来全是RT，POJ出故障了！！！！

Card Trick

Time Limit:1000MS		Memory Limit:65536K

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 thenth and last card turns out to be thenof 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 integern.

Output

For each test case, output a line with the correct permutation of the values 1 ton, 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

/* Author : yan
 * Question : POJ 3032 Card Trick
 * Data && Time : Wednesday, January 05 2011 10:50 PM
*/
#include<stdio.h>
typedef struct _node
{
	int value;
	int pos;
};
struct _node qu[150];
struct _node stack[14];
int cnt_stack;

int head,rear;

int enqueue(struct _node node)
{
	qu[rear++]=node;
	return rear;
}
struct _node dequeue()
{
	return qu[head++];
}
int cmp(const void *a,const void *b)
{
	return (*(struct _node *)a).pos - (*(struct _node *)b).pos;
}
int main()
{
	//freopen("input","r",stdin);
	int n;
	int i,j;
	int test;
	struct _node cache;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d",&n);
		cnt_stack=0;
		head=rear=0;
		for(i=0;i<n;i++)
		{
			struct _node node;
			node.pos=i;
			enqueue(node);
		}
		for(i=1;i<=n;i++)
		{
			for(j=0;j<i;j++)
			{
				cache=dequeue();
				enqueue(cache);
			}
			cache=dequeue();
			cache.value=i;
			stack[cnt_stack++]=cache;
		}

		qsort(stack,cnt_stack,sizeof(stack[0]),cmp);
		for(i=0;i<cnt_stack;i++)
		{
			printf("%d ",stack[i].value);
		}
		printf("/n");
	}
	return 0;
}