(compareTo) How Many Fibs hdu1316 && ZOJ1962

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7007    Accepted Submission(s): 2761

Problem Description

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].

Input

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 <= 10^100. The numbers a and b are given with no superfluous leading zeros.

Output

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

Sample Input

10 100

1234567890 9876543210

0 0

Sample Output

5

4

用Java，然后利用compareTo来判断大小。

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        BigInteger a,b;
        while(in.hasNextBigInteger()) {
            a=in.nextBigInteger();
            b=in.nextBigInteger();
            if(a.equals(BigInteger.valueOf(0))&&b.equals(BigInteger.valueOf(0)))
                break;                 //也可以用compareTo。
            System.out.println(Fib(a,b));
        }
    }
     public static int Fib(BigInteger a,BigInteger b)
        {
            int sum = 0;
            BigInteger f = new BigInteger("1");
            BigInteger s = new BigInteger("2");
            BigInteger tmp;
            while(true)
            {
                if(f.compareTo(b)>0)
                    break;
                if(f.compareTo(a) >= 0)
                    sum++;
                tmp = f;                      //可以用迭代的方法求斐波那契数列。
                f = s;
                s = s.add(tmp);
            }
            return sum;
        }
}

原文地址：https://www.cnblogs.com/Weixu-Liu/p/9166478.html