HDOJ/HDU 1250 Hat's Fibonacci(大数~斐波拉契)

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<f.length;i++){ f[i]=f[i-1].add(f[i-2]).add(f[i-3]).add(f[i-4]); } }
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/51472142