hdu 1250 Hat's Fibonacci （Java大数计算）

最新推荐文章于 2021-03-11 02:36:45 发布

慧丫儿

最新推荐文章于 2021-03-11 02:36:45 发布

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本文链接：https://blog.youkuaiyun.com/qq_42819248/article/details/83998483

本文介绍了一种使用Java实现四阶斐波那契数列的方法，通过预计算并存储前10000个数值，使得能够快速查询任意指定位置的数列值。适用于处理大整数运算，确保了高效率与准确性。

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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.

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
        Scanner in = new Scanner(System.in);
        BigInteger[] f = new BigInteger[10005];
        f[1] = f[2] = f[3] = f[4] = BigInteger.valueOf(1); //赋值格式
        int i;
        for(i = 5; i <= 10000; i++)
        {
        	f[i] = f[i-1].add(f[i-2]);
        	f[i] = f[i].add(f[i-3]);
        	f[i] = f[i].add(f[i-4]);
        }
        while(in.hasNext())
        {
        	
        	int n = in.nextInt(); //当n超过30时，就需要用大数类
        	
        	System.out.println(f[n]);
        }
	}

}