nyoj714 Card Trick(第六届河南省程序设计大赛)

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描述

The magician shuffles a small pack of cards, holds it face down and performs the following procedure:

1. 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.
2. 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.
3. Three cards are moved one at a time…
4. 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.

输入

On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 10 Each case consists of one line containing the integer n. 1 ≤ n ≤ 13

输出

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…

样例输入

2
4
5

样例输出

2 1 4 3
3 1 4 5 2

来源

第六届河南省程序设计大赛

上传者

ACM_赵铭浩

题意：给你n张牌，分别是1-n

 第一次把最上面1张放到最下面，亮出当前最上面的牌 使其为1

 第二次把最上面2张放到最下面，亮出当前最上面的牌 使其为2

 第三次把最上面3张放到最下面，亮出当前最上面的牌 使其为3

..............

模拟一下过程就好了 当然如果你对时间更苛刻 由于最多只有13张牌 可以把结果存在一个数组  

#include <stdio.h>
#include <queue>
using namespace std;
int main()
{
	int ncase;
	int order[13];
	scanf("%d",&ncase);
	while(ncase--)
	{
		int n;
		scanf("%d",&n);
		queue<int>s;
		for(int i=0;i<n;i++)
		s.push(i);
		int k=1;
		while(!s.empty())
		{
			for(int i=0;i<k;i++)
			{
				int x=s.front();s.pop();
				s.push(x);
			}
			int y=s.front();s.pop();
			order[y]=k++;
		}
		printf("%d",order[0]);
		for(int i=1;i<n;i++)
		printf(" %d",order[i]);
		printf("\n");
	}
	return 0;
}

nyoj714 Card Trick(第六届河南省程序设计大赛)

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原文地址：http://blog.csdn.net/su20145104009/article/details/51272719