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nyoj714 Card Trick (The 6th Henan Province Programming Contest)

2022-08-08 19:07:00 【51CTO】


Card Trick

1000 ms  |  内存限制： 65535

3


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 thenof 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.


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
      
      

      
      
    
       
       
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    • 
       
       
  • 2.
    • 
       
       
  • 3.
    •

样例输出 

      
      

       
       2 1 4 3
       
       
3 1 4 5 2
      
      

      
      
    
       
       
  • 1.
    • 
       
       
  • 2.
    •

来源 ​ ​第六届河南省程序设计大赛​​

上传者

​ ​ACM_赵铭浩​​


题意：给你n张牌,分别是1-n

 First put on top1Zhang put it at the bottom,Reveal the current top card 使其为1

 The second time put on top2Zhang put it at the bottom,Reveal the current top card 使其为2

 The third time put on top3Zhang put it at the bottom,Reveal the current top card 使其为3

..............

Just simulate the process Of course if you are more critical of time 由于最多只有13张牌 The result can be stored in an array  


      
      

       
       #include <stdio.h>
       
       

       
       #include <queue>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       int 
       
       main()
       
       
{
       
       
  
       
       int 
       
       ncase;
       
       
  
       
       int 
       
       order[
       
       13];
       
       
  
       
       scanf(
       
       "%d",
       
       &
       
       ncase);
       
       
  
       
       while(
       
       ncase
       
       --)
       
       
  {
       
       
    
       
       int 
       
       n;
       
       
    
       
       scanf(
       
       "%d",
       
       &
       
       n);
       
       
    
       
       queue
       
       <
       
       int
       
       >
       
       s;
       
       
    
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       n;
       
       i
       
       ++)
       
       
    
       
       s.
       
       push(
       
       i);
       
       
    
       
       int 
       
       k
       
       =
       
       1;
       
       
    
       
       while(
       
       !
       
       s.
       
       empty())
       
       
    {
       
       
      
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       k;
       
       i
       
       ++)
       
       
      {
       
       
        
       
       int 
       
       x
       
       =
       
       s.
       
       front();
       
       s.
       
       pop();
       
       
        
       
       s.
       
       push(
       
       x);
       
       
      }
       
       
      
       
       int 
       
       y
       
       =
       
       s.
       
       front();
       
       s.
       
       pop();
       
       
      
       
       order[
       
       y]
       
       =
       
       k
       
       ++;
       
       
    }
       
       
    
       
       printf(
       
       "%d",
       
       order[
       
       0]);
       
       
    
       
       for(
       
       int 
       
       i
       
       =
       
       1;
       
       i
       
       <
       
       n;
       
       i
       
       ++)
       
       
    
       
       printf(
       
       " %d",
       
       order[
       
       i]);
       
       
    
       
       printf(
       
       "\n");
       
       
  }
       
       
  
       
       return 
       
       0;
       
       
}
      
      

      
      
    
       
       
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