NYOJ714 Card Trick 【队列模拟】

最新推荐文章于 2019-04-29 23:08:13 发布

长风Qiu

最新推荐文章于 2019-04-29 23:08:13 发布

阅读量1.3k

点赞数

CC 4.0 BY-SA版权

文章标签： NYOJ714

本文链接：https://blog.youkuaiyun.com/chang_mu/article/details/25558687

探讨了一种独特的纸牌排列算法，该算法通过一系列特定的操作确保每张牌按顺序出现，从A到K，揭示了背后的数学原理及实现方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Card Trick

时间限制：1000 ms | 内存限制：65535 KB

难度：3

描述

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.

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

样例输出

2 1 4 3

3 1 4 5 2

 /*
 **题意是给定一个1到n的某种初始序列，然后将列首数放到序列的末尾（操作一次），将新的队首元素放到
 **桌面，再将队首的两个数放到末尾（操作两次），新树放到桌面，以此类推。最终得到的数列是1到n的递
 **增序列。求初始序列。
 **这是一道模拟题，可以逆推：由于最终序列一定是1，2，3...n，所以最后一个摆在桌面上的数一定是n，
 **且n被“操作”了n次，倒数第二个被放到桌面的数一定是n-1且n-1一定和n一起被“操作”了n-1次，以此类推。
 **可以倒着模拟，即先将最终数列放到数组里，再从第n个数开始模拟，即将第n个数向前“操作”n次，实际上
 **位置没变，然后将第n-1个数加入到“操作”中且与第n个数一起向前“操作”n-1次，以此类推，最终得到的数
 **列即是初始序列。
 */
#include <stdio.h>
#include <string.h>
int a[15];

void move(int bot, int top, int n){
	while(n--){
		int t = a[top], tt;
		for(int i = bot; i <= top; ++i){
			tt = a[i];
			a[i] = t;
			t = tt;
		}
	}
}

int main(){
	int t, n, bot, top;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		for(int i = 1; i <= n; ++i) a[i] = i;
		top = bot = n;
		for(int i = n; i >= 1; --i){
			move(bot, top, i % (top - bot + 1));
			--bot;
		}
		for(int i = 1; i <= n; ++i)
			printf("%d ", a[i]);
		printf("\n");
	}
	return 0;
}