HDU2319— Card Trick

最近才看的STL—vector,是以就練練手手，寫了一道類似的題。 ## 題目：J - Card Trick

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

2

4

5

Sample Output

2 1 4 3

3 1 4 5 2

解題思路：

利用尾插法先把第一個數n輸入進去，當輸出的數組的長度不滿足n時執行一下循環：把第n-i個數放在頭部，把最後一個數放在頭部，删除尾部的數，每一次循環讓i++；

注意：最後一個尾數沒有空格。

程式代碼：

#include<vector>
#include<iostream>
using namespace std;
int main()
{
	vector<int> v;
	int i,n,m,k,b;
	cin>>k;
	for(int j=1;j<=k;j++)
	{
		cin>>n;
		v.push_back(n);
		i=1;
		while(v.size()!=n)
		{
			v.insert(v.begin(),n-i);
			m=n-i;
			while(m--)
			{
				v.insert(v.begin(),*(v.end()-1));
				v.erase(v.end()-1);
			}
			i++;	
		}
		for(i=0;i<n-1;i++)
			cout<<v[i]<<" ";
		cout<<v[n-1];
		cout<<endl;
		v.clear();	
	}
	return 0;
} 
 
           

vector用法不懂的可以參照上一篇部落格