NYOJ 714 Card Trick(队列queue）

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 thenth 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

题意：

第一次把牌最上面的一个放到牌的最下面，然后此时最上面的牌为黑桃一并删除，第二次把最上面的两个放到最下面，此时最上面的为黑桃二并删除，以此类推n次；

思路：首先建一个队列存一到n，将前1个放到队尾，此时的队顶元素既是找到的黑桃被删除的数，以此类推就可以找到原序列

代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;int c[20];int main(){    int t;    scanf("%d",&t);    while(t--)    {        queue<int>q;        int n;        scanf("%d",&n);        for(int i=1; i<=n; i++)        {            q.push(i);        }        int k=1;        for(int i=1; i<=n; i++)        {            for(int j=1; j<=i; j++)            {                int ans=q.front();                q.pop();                q.push(ans);            }            int dir=q.front();            c[dir]=k++;//k从2开始            q.pop();            //printf("%d %d\n",dir,k);        }        for(int i=1; i<=n; i++)        {            if(i!=n)            {                printf("%d ",c[i]);            }            else            {                printf("%d\n",c[i]);            }        }    }    return 0;}