河南省第六届程序设计大赛--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

2

4

5

样例输出

2 1 4 3

3 1 4 5 2

#include <iostream>#include <queue>using std::endl;using std::cout;using std::cin;using std::queue;int main(){//data 用来保存最后得到的序列int data[13];queue<int> q;int K;cin >> K;while(K--){int m;cin >> m;for(int i=0; i<m; ++i){q.push(i);}for(int i=0;i<m;++i){//cnt用来记录每次需要移动的数量int cnt = i+1;while(cnt--){//temp用来记录每次队列中队首的元素int temp = q.front();q.pop();q.push(temp);}//队列中的元素进行移动过后队首的元素就是排列中的一个元素的下标data[q.front()] = i+1;//此时得到排列中的一个坐标后就将其删去q.pop();}//输出最后的排列for(int i=0; i<m; ++i){cout << data[i] << " ";}cout << endl;}system("pause");return 0;}