Tower of Hanoi 2 (function)

describe

the The classical tower of Hanoi problem often exists as a classic example of recursion . Maybe someone doesn't know the story of the tower of Hanoi problem . The tower of Hanoi comes from a story in Indian legend , God made three diamond pillars when he created the world Son , On a column, they are stacked in order of size from bottom to top 64 A golden disk . God ordered Brahman to rearrange the disc on another post in order of size from below . And stipulate , You can't put on a small disc Large disc , Only one disc can be moved between the three pillars at a time . There is a prophecy that , When this is done, the universe will be destroyed in a flash . Some people believe that Brahmans are still moving the disc all the time . Okay , When However, this legend is not believable , Nowadays, Hanoi Tower exists more as a toy .Gardon I received a Hanoi Tower toy as a birthday gift .

Gardon He is afraid of trouble （ Okay , Is a lazy person ）, It is obvious that 64 It's difficult to move the plates one by one until all the plates reach the third pillar , therefore Gardon Decide to make a Minor malpractice , He found another exactly the same post , Move all the dishes to the third pillar faster through this pillar . The following question is ： When Gardon Used in a game N null when , How many times does he need to move them to the third pillar ？ Obviously , When there is no fourth pillar , The solution to the problem is 2^N-1, But now with the help of this pillar , How much should it be ？

Input
Contains multiple sets of data , One line per data , It's the number of plates N(1<=N<=64).

Output
For each set of data , Output a number , The minimum number of moves required to reach the target .

sample input 1

1
3
12

sample output 1

1
5
81

#include<stdio.h>
#include<math.h> 

int main(){
    
	int a[65];
	int i,j,temp,min;
	a[1] = 1;
	a[2] = 3;
	a[3] = 5;
	for(i=4;i<65;i++){
    
		min = 100000;
		for(j=1;j<i;j++){
    
			if((a[j]*2+pow(2,i-j)-1)<min){
    
				min = a[j]*2+pow(2,i-j)-1;
			}
		}
		a[i] = min;
	}
	int n;
	while(scanf("%d",&n)!=EOF){
    
      	if(n==0) break;
		printf("%d\n",a[n]);
	}
}