We want to represent positive integer n as a sum of addends, where each addends is an integer number containing only 1s. For example, we can represent 121 as 121=111+11+–1. Help us to find the least number of digits 1 in such sum. Input: The first line of the input contains integer n (1 ≤ n < 1015). Output: Print expected minimal number of digits 1. (ex. input:121, output:6)

C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN:9781337102087
Author:D. S. Malik
Publisher:D. S. Malik
Chapter5: Control Structures Ii (repetition)
Section: Chapter Questions
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I wrote the following program for this problem and it works, but is too slow on big numbers. How could i make it faster?

Problem:

We want to represent positive integer n as a sum of addends, where each addends is an integer number containing only 1s. For example, we can represent 121 as 121=111+11+–1. Help us to find the least number of digits 1 in such sum.

Input: The first line of the input contains integer n (1 ≤ n < 1015).
Output: Print expected minimal number of digits 1.
(ex. input:121, output:6)
 
My solution:
 

import java.util.Scanner;

public class Main {

staticlong[] en=newlong[]{1,11,111,1111,11111,111111,1111111,11111111,111111111,1111111111,

11111111111L,111111111111L,1111111111111L,11111111111111L,111111111111111L,1111111111111111L};

 

public static void main(String[] args){

Scanner scanner =new Scanner(System.in);

System.out.println(rekisto(scanner.nextLong()));

}

 

static long rekisto(long s){

   if(s==0){

       return 0;

    }

int dolzina =String.valueOf(s).length();

    if(Math.abs((en[dolzina])-s)<s){

       long n2 = rekisto(Math.abs((en[dolzina])-s))+(dolzina+1);

       return (Math.min((rekisto(Math.abs(s-en[dolzina-1]))+ dolzina),n2));

    }

return rekisto(Math.abs(s-en[dolzina-1]))+ dolzina;

}

}

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