How many Fibs?【sudt 2321】【大数的加法及其比较】

How many Fibs?

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

Recall the definition of the Fibonacci numbers:

f₁ := 1 
f₂ := 2 
f_n := 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 aand b are given with no superfluous leading zeros.

输出

 

示例输入

10 100
1234567890 9876543210
0 0

示例输出

5
4

提示

 

来源

2000/2001 University of Ulm Local Contest

示例程序

题目大意：

输入两个整数，求这两个整数之间的斐波纳契数的个数，即求[a,b]之间斐波那契数的个数，输入以0 0结束（a,b两个数可以到10^100次幂）

代码：

很多大数的问题用java做的话一定比c语言或者c++语言做起来简单~

import java.io.*;
import java.math.*;
import java.util.*;
public class Main
{
    public static void main(String args[])
    {
        Scanner scn=new Scanner(System.in);
        while(true)
        {
            BigInteger a=scn.nextBigInteger();
            BigInteger b=scn.nextBigInteger();
            if(a.equals(new BigInteger("0"))&&b.equals(new BigInteger("0")))
            {
                break;
            }
            BigInteger f[]=new BigInteger[20000];
            f[1]=new BigInteger("1");
            f[2]=new BigInteger("2");
            int i;
            for(i=3;i<=600;i++)
            {
                int temp=i;
                f[i]=f[temp-1].add(f[temp-2]);
            }
            int count=0;
            for(i=1;i<=600;i++)
            {
                if(f[i].compareTo(a)==0)
                {
                    count=1;
                }
                else if(f[i].compareTo(a)>0&&f[i].compareTo(b)<=0)
                {
                    count++;
                }
                else if(f[i].compareTo(b)>0)
                {
                    break;
                }
            }
            System.out.println(count);
        }
    }
}