标签：big case public can AC several 斐波那契数列 equals out
How Many Fibs?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7007 Accepted Submission(s): 2761
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
用Java，然后利用compareTo来判断大小。
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
BigInteger a,b;
while(in.hasNextBigInteger()) {
a=in.nextBigInteger();
b=in.nextBigInteger();
if(a.equals(BigInteger.valueOf(0))&&b.equals(BigInteger.valueOf(0)))
break; //也可以用compareTo。
System.out.println(Fib(a,b));
}
}
public static int Fib(BigInteger a,BigInteger b)
{
int sum = 0;
BigInteger f = new BigInteger("1");
BigInteger s = new BigInteger("2");
BigInteger tmp;
while(true)
{
if(f.compareTo(b)>0)
break;
if(f.compareTo(a) >= 0)
sum++;
tmp = f; //可以用迭代的方法求斐波那契数列。
f = s;
s = s.add(tmp);
}
return sum;
}
}
(compareTo) How Many Fibs hdu1316 && ZOJ1962
标签：big case public can AC several 斐波那契数列 equals out
原文地址：https://www.cnblogs.com/Weixu-Liu/p/9166478.html
