HDU 1316 How Many Fibs? (大Fib数,还是Java大法好)

Posted 2020-07-07 AC_Dreameng

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5822    Accepted Submission(s): 2267

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

 

Source

University of Ulm Local Contest 2000

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1316

题意:求指定区间里有多少个Fib数.

果断打表,将所有的Fib数按顺寻存在数组中,但是数组长度是多大呢?题目中只说了Fib数的的区间上限是10^100,期间有多少个Fib数并不知道,所以首先得确定数组大小!

AC代码:

import java.math.BigDecimal;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		final int MAXN = 1000+5;
		BigDecimal [] a = new BigDecimal[MAXN];
		a[1]=new BigDecimal(1);
		a[2]=new BigDecimal(2);
		for (int i = 3; i <MAXN; i++) {
			a[i] = a[i-1].add(a[i-2]);
		}
		//System.out.println(a[MAXN-1]);
		//System.out.println(a[MAXN-1].toString().length());
		while(sc.hasNext()){
			BigDecimal start = sc.nextBigDecimal();
			BigDecimal end = sc.nextBigDecimal();
			BigDecimal zero = BigDecimal.ZERO;
			if(start.compareTo(zero)==0&&end.compareTo(zero)==0){
				break;
			}
			int sum = 0;
			for (int i = 1; i < MAXN; i++) {
				if(a[i].compareTo(start)<0){
					continue;
				}else if(a[i].compareTo(end)>0){
					break;
				}else{
					sum++;
				}
			}
			System.out.println(sum);
		}

	}

}