hdu 1316(大整数)

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5954    Accepted Submission(s): 2331

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

 

 

Java大整数

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
         Scanner sc = new Scanner(System.in);
         BigInteger[] fib = new BigInteger[];
         BigInteger MAX = BigInteger.valueOf().pow();
         fib[] = BigInteger.valueOf();
         fib[] = BigInteger.valueOf();
         for(int i=;i<=;i++){
             fib[i] = fib[i-].add(fib[i-]);
            // System.out.println(fib[i]);
         }
         while(sc.hasNext()){
             String str1 = sc.next();
             String str2 = sc.next();
             if(str1.equals("")&&str2.equals("")) break;
             BigInteger a = new BigInteger(str1);
             BigInteger b = new BigInteger(str2);
             int start=,end = ;
             for(int i=;i<=;i++){
                 if(fib[i].compareTo(a)>=) {
                     start = i;
                     break;
                 }
             }
             for(int i=;i<=;i++){
                 if(fib[i].compareTo(b)==) {
                     end = i;
                     break;
                 }
                 if(fib[i].compareTo(b)>){
                     end = i-;
                     break;
                 }
             }
             //System.out.println(start+" "+end);
             System.out.println(end-start+);
         }
     }
}