HackerRank Fibonacci Modified

原题网址：https://www.hackerrank.com/challenges/fibonacci-modified

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 Brespectively,
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 B respectively.

Note

Some output may even exceed the range of 64 bit integer.

Sample Input

0 1 5

Sample Output

5

Explanation

The first two terms of the series are 0 and 1. The fifth term is 5. How we arrive at the fifth term, is explained step by step in the introductory sections.
方法：动态规划。

import java.io.*;
import java.util.*;
import java.math.BigInteger;

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 scanner = new Scanner(System.in);
int a = scanner.nextInt();
int b = scanner.nextInt();
int n = scanner.nextInt();
BigInteger[] nums = new BigInteger
;
nums[0] = new BigInteger(Integer.toString(a), 10);
nums[1] = new BigInteger(Integer.toString(b), 10);
for(int i = 0; i < n - 2; i ++) {
nums[i + 2] = nums[i].add(nums[i + 1].multiply(nums[i + 1]));
}
System.out.println(nums[n - 1].toString());
}
}