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);
         }
     }
 }