Problem Statement
A series is defined in the following manner:
Given the nth and
(n+1)th terms,
the (n+2)th can
be computed by the following relation
Tn+2 =
(Tn+1)2 +
Tn
So, if the first two terms of the series are 0 and 1:
the third term = 12 +
0 = 1
fourth term = 12 +
1 = 2
fifth term = 22 +
1 = 5
... And so on.
Given three integers A, B and N, such that the first two terms of the series (1st and 2nd terms) are A and B respectively, compute the Nth term of the series.
Input Format
You are given three space separated integers A, B and N on one line.
Input Constraints
0 <= A,B <= 2
3 <= N <= 20
Output Format
One integer.
This integer is the Nth term
of the given series when the first two terms are A and Brespectively.
Note
- Some output may even exceed the range of 64 bit integer.
Sample Input
0 1 5
Sample Output
5
import java.math.BigInteger;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner cin=new Scanner(System.in);
int a=cin.nextInt();
int b=cin.nextInt();
int n=cin.nextInt();
int k=2;
BigInteger i = BigInteger.valueOf(a);
BigInteger j = BigInteger.valueOf(b);
BigInteger res ;
while(n!=k){
k++;
res=i.add(j.multiply(j));
i=j;
j=res;
}
System.out.println(j);
}
}