Hdoj HDU 2064 tower iii acm 2064 in HDU

// Original miyu, please note: Reprinted from __________ White House

Question address:
Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2064
Description:
Tower III
About the end of the 19th century, I sold an intellectual toy in a store in ozhou with three poles on a copper plate, on the leftmost bar, the Tower consists of 64 disks in ascending order. The purpose is to move all the disks on the leftmost bar to the right bar, with the condition that only one disk can be moved at a time and the tray cannot be placed on a small disk.
Now we can change the gameplay. We cannot directly move from the leftmost (rightmost) side to the rightmost (leftmost) side (each move must be moved to or from the middle ), you cannot place the dashboard on the lower disk.
Daisy has already done the original tower problem and Tower II, but when she encountered this problem, she thought for a long time and couldn't solve it. Please help her now. Now there are n discs. How many times does she move these discs from the leftmost to the rightmost?

Tower of Hanoi is a classic recursive instance,If the rule is not so abnormal, allow direct jump from 1 to 3, then n disks need at least 2n-1.

Some new rules are added here. We can analyze them as follows:NDisk (s) from1Move3:

Step 1: initial status:

Step 1: Move n-1-1 disks above to the 2nd ROD:

Step 1: Move the nth disk from 1 to 2:

Step 1: Move the first n-1 from 3 to 1 to give way to the first disk:

Step 1: Move the nth disk from 2 to 3:

Step 1: Move the first n-1 from 3 to complete the movement:

We set F (n) to the number of steps required to move n disks from 1 to 3, which is also the number of steps required to move from 3 to 1.

From the figure above, we can see that it takes three steps to move the nth disk from 1 to 3:

1) Move the first n-1 from 1. 3.

2) the nth disk must go through two steps from 1-> 2-> 3.

3.) n-1-1 Disk needs to be 3-> 1 (this is to give way to the n-disk), and then 1-> 3.

When F (n) = 3 × F (n-1) + 2;

F (1) = 2;

In this way, we get the recursive formula for this question. Of course, we can further optimize it. The optimization method is as follows:

F (n) = 3 × F (n-1) + 2
F (1) = 2
=>
F (n) + 1 = 3 × [F (n-1) + 1]
F (1) + 1 = 2 + 1 = 3
=>
F (n) + 1 = 3n
=>
F (n) = 3N-1
Last pastedCode:
 

 // Original miyu, please note: Reprinted from __________ White House

# Include  <  Iostream  >  
# Include  < Cmath  >  
  Using     Namespace  STD;
  Long     Long  MYPOW (  Int  N,  Int  E)
{
 Long     Long  MLT  =     1  ;
  For  (  Int  I  =     1  ; I <=  E;  ++  I)
{
MLT  * =  N;
}
  Return  MlT;
}
  Int  Main ()
{
  Int  N;
 While  (CIN  >  N)
{
Cout  <  MYPOW (  3  , N)  -     1     < Endl;
}
  Return     0  ;
}