nyoj 714 Card Trick

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:

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

245

样例输出

2 1 4 33 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

代码：

#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>using namespace std;int a[20];int top,bottom;void move(int bottom,int top,int n){while(n--){int tt=a[top],t;for(int i=bottom;i<=top;i++){t=a[i];a[i]=tt;tt=t;}}}int main(){int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);for(int i=1;i<=n;i++){a[i]=i;}top=bottom=n;//保存数组的起点和终点 for(int i=n;i>=1;i--)//i代表移动的次数 {//因为i/(top-bottom+1)*(top-bottom+1)次的移动会使数组保持不变 move(bottom,top,i%(top-bottom+1));//数组的起点，终点，移动的次数 bottom--;//每次加一个数 }printf("%d",a[1]);//输出最初的数组 for(int i=2;i<=n;i++){printf(" %d",a[i]);}printf("\n");}return 0;}