How many Fibs?

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 := fn-1 + fn-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);
        }
    }
  
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