How many Fibs?

最新推荐文章于 2021-04-18 09:15:45 发布

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最新推荐文章于 2021-04-18 09:15:45 发布

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分类专栏： java大数

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本文介绍了一个算法，用于计算给定范围内斐波那契数的数量。通过生成斐波那契数列并检查其是否在指定范围内，算法能够高效地找到答案。

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题目描述

Recall the definition of the Fibonacci numbers:

f1 := 1 

f2 := 2 

fn := f_n-1 + f_n-2     (n>=3)

Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a,b].

输入

The input contains several test cases. Each test case consists of two non-negative integer numbers a and b. Input is terminated by a=b=0. Otherwise, a<=b<=10¹⁰⁰. The numbers a and b are given with no superfluous leading zeros.

输出

For each test case output on a single line the number of Fibonacci numbers fi with a<=fi<=b.

样例输入

10 100
1234567890 9876543210
0 0

样例输出

5
4

import java.math.BigInteger; 
import java.util.Scanner; 
public class Main { 
    public static void main(String[] args) { 
        BigInteger f[] = new BigInteger[505]; 
        f[1] = BigInteger.ONE; 
        f[2] = BigInteger.valueOf(2); 
        for (int i = 3; i < 505; ++i) 
            f[i] = f[i - 1].add(f[i - 2]); 
        Scanner scanner = new Scanner(System.in); 
        while (scanner.hasNext()) { 
            BigInteger nBigInteger = scanner.nextBigInteger(); 
            BigInteger mBigInteger = scanner.nextBigInteger(); 
            if (nBigInteger.equals(BigInteger.ZERO) 
                    && mBigInteger.equals(BigInteger.ZERO)) 
                break; 
            int c = 0; 
            for (int i = 1; i < 505; ++i) { 
                if (f[i].compareTo(mBigInteger) > 0) 
                    break; 
                if (f[i].compareTo(nBigInteger) >= 0
                        && f[i].compareTo(mBigInteger) <= 0) 
                    ++c; 
            } 
            System.out.println(c); 
        } 
    } 
  
}