hdu--1316--How Many Fibs?(java大数)
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How Many Fibs?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6371 Accepted Submission(s): 2517
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].
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
1 import java.util.Scanner; 2 import java.math.*; 3 4 /** 5 * @author 日天大帝 6 * 第480个斐波数是101位,打个表开480够用 7 */ 8 public class Main { 9 public static void main(String[] args) { 10 Scanner cin = new Scanner(System.in); 11 final int MAX = 480; 12 BigInteger arr[] = new BigInteger[MAX]; 13 arr[0] = BigInteger.ONE; 14 arr[1] = BigInteger.valueOf(2); 15 for(int i=2; i<MAX ; ++i){ 16 arr[i] = arr[i-1].add(arr[i-2]); 17 } 18 while(cin.hasNext()){ 19 BigInteger a = cin.nextBigInteger(); 20 BigInteger b = cin.nextBigInteger(); 21 if(a.compareTo(b) == 0 && a.compareTo(BigInteger.ZERO) == 0)break; 22 int ct = 0; 23 for(int i=0;; ++i){ 24 if(arr[i].compareTo(b) > 0)break; 25 if(arr[i].compareTo(a) >= 0)ct++; 26 } 27 System.out.println(ct); 28 } 29 } 30 }
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