HDU：2319 Card Trick

Card Trick

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 273    Accepted Submission(s): 132

Problem Description

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.

 

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 integer n.

 

Output

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…

 

Sample Input

245

 

Sample Output

2 1 4 33 1 4 5 2题目意思：一副牌，有1~13，这些牌，按题目所给规则翻拍后，牌被取走的顺序为1~n例如：2 1 4 3，先将前1张牌放到最后面，序列为1 4 3 2，然后取走1，剩余 4 3 2，然后在挪动两次牌，顺序变为2 3 4，然后取走2，剩余3 4，然后挪动3次牌，变为3 4，拿走3，剩余4，牌被拿走的顺序为1 2 3 4，让求该顺序所对应的牌的原始序列。题目代码：#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>using namespace std;int main(){    int m,n;//m组测试数据    int a[15],b[15];    cin>>m;    while(m--)    {        cin>>n;//测试数据        queue<int>qu;//定义一个队列        for(int i = 1; i <= n; i++)//将1~n入队，i代表牌的位置        {            qu.push(i);        }        for(int i = 1; i <= n; i++)//求n张牌的位置        {            int temp = i;//翻牌次数            while(temp--)            {                int x = qu.front();                qu.pop();                qu.push(x);            }            a[i] = qu.front();//保留队首元素的值，即为i在牌中的初始次序            qu.pop();        }        for(int i = 1; i <= n; i++)        {            b[a[i]] = i;//a[i]为i在牌中位置        }        cout<<b[1];        for(int i = 2; i <= n; i++)        {            cout<<" "<<b[i];        }        cout<<endl;    }    return 0;}