http://acm.hdu.edu.cn/showproblem.php?pid=1316
How Many Fibs?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2354 Accepted Submission(s): 955
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
Source
Recommend
Eddy
View Code
1 import java.math.BigInteger; 2 import java.util.Scanner; 3 4 5 public class Main { 6 public static void main(String[] args) { 7 BigInteger f[]=new BigInteger[1010]; 8 int a[]=new int[1010]; 9 BigInteger aa,bb; 10 Scanner cinScanner=new Scanner(System.in); 11 f[1]=BigInteger.ONE; 12 f[2]=BigInteger.valueOf(2); 13 for(int i=3;i<1001;i++) 14 f[i]=f[i-1].add(f[i-2]); 15 int n; 16 while(cinScanner.hasNext()) 17 { 18 aa=cinScanner.nextBigInteger(); 19 bb=cinScanner.nextBigInteger(); 20 if(aa.compareTo(BigInteger.ZERO)==0&&bb.compareTo(BigInteger.ZERO)==0) 21 break; 22 if(aa.compareTo(BigInteger.ONE)==0&&bb.compareTo(BigInteger.ONE)==0) 23 { 24 System.out.println("1"); 25 continue; 26 } 27 int j=1; 28 while(f[j].compareTo(aa)<0) 29 { 30 j++; 31 } 32 //System.out.println(j); 33 int count=0; 34 while((f[j]).compareTo(bb)<=0) 35 { 36 count++; 37 j++; 38 } 39 System.out.println(count); 40 41 42 } 43 44 45 } 46 47 }