Soluție HackerRank pentru Little Gaurav and Sequence, subdomeniul Algebra, în Java 8. Include cerința formatată, exemple, explicația pașilor și cod sursă.
- Problemă: Little Gaurav and Sequence
- Domeniu: Algebra
- Limbaj: Java 8
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
Cerință
Little Gaurav is very fond of numbers and sequences. One day his teacher tells him to find a strange sequence.
S = ∑_i=0, 2^i≤ n^∞ ∑_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 ≤ T ≤ 1000
1 ≤ N ≤ 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 sursă
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);
}
}
}