本文共 1414 字，大约阅读时间需要 4 分钟。

Problem Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1. F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4) Your task is to take a number as input, and print that Fibonacci number.

Input

Each line will contain an integers. Process to end of file.

Output

For each case, output the result in a line.

Sample Input

100

Sample Output

4203968145672990846840663646

Note:

No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

就是根据这个公式：

F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)

输入一个n，输出f(n)的值。

注意，这是大数~答案的位数高达2005位~~~

再一次体会Java大数的强大吧~

import java.math.BigInteger;import java.util.Scanner;public class Main {    static BigInteger f[] = new BigInteger[7045];    public static void main(String[] args) {        dabiao();        Scanner sc = new Scanner(System.in);        while(sc.hasNext()){            int n =sc.nextInt();            System.out.println(f[n]);            //System.out.println("---------");            //System.out.println(f[n].toString().length());            //开数组~看开到多少位的时候，位数大于2005        }    }    private static void dabiao() {        f[1]=new BigInteger("1");        f[2]=new BigInteger("1");        f[3]=new BigInteger("1");        f[4]=new BigInteger("1");        for(int i=5;i

转载地址：https://chenhx.blog.csdn.net/article/details/51472142 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！