HackerRank Algebra – Little Gaurav and Sequence

Challenge: Little Gaurav and Sequence

Subdomeniu: Algebra (algebra)

Scor cont: 20.0 / 20

Submission status: Accepted

Submission score: 1.0

Submission ID: 464721314

Limbaj: java8

Link challenge: https://www.hackerrank.com/challenges/little-gaurav-and-sequence/problem

Cerinta

Little Gaurav is very fond of numbers and sequences. One day his teacher tells him to find a strange sequence.    

$$S = \sum_{i=0, 2^{i}\leq n}^\infty     \sum_{j=0}^n 2^{2^{i}+2j}$$

Since this sequence looks a bit difficult, the teacher tells him to find the last digit of $S$.  

Little Gaurav is confused because he cannot solve the problem and leaves this problem to the worthy programmers of the world. Help little Gaurav in finding the solution.   

**Input Format**  
The first line contains $T$, the number of test cases.  
$T$ lines follow, each line containing an integer, $N$.   

**Output Format**  
For each testcase, print the last digit of $S$ in one line.    

**Constraints**   
$1 \le T \le 1000$  
$1 \le N \le 10^{15}$  

**Sample Input**     

    3
    1
    2
    3
    
**Sample Output**   

    0
    6
    0

**Explanation**  
For n=1, only i=0 is valid. So S is $2^{2^{0}+0} + 2^{2^{0}+2} = 10$. Hence last digit of S is 0.   
For n=2, only i=0 and 1 are valid. So S is    
S1(for i=0) is $2^{2^{0}+0} + 2^{2^{0}+2} + 2^{2^{0}+4}= 42$.    
S2(for i=1) is $2^{2^{1}+0} + 2^{2^{1}+2} + 2^{2^{1}+4}= 84$.   
So last digit of S is 6.

Cod sursa

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int t=in.nextInt();
        for(int a0=0;a0<t;a0++){
            long n=in.nextLong();
            double x=Math.log(n)/Math.log(2);
            int y=(int)x+1;
            int s=0;
            if(y==1)
                s=2;
            else
                s=(y-1)*6;
            int a=s%10;
            int b=5;
            if(n%2==0)
                b=1;
            int c=(a*b)%10;
            System.out.println(c);
        }
        
    }
}