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[n];        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());    }}