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2013
11-11

POJ 2413 How many Fibs? [解题报告] Java

How many Fibs?

问题描述 :

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].

输入:

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<=10100. The numbers a and b are given with no superfluous leading zeros.

输出:

For each test case output on a single line the number of Fibonacci numbers fi with a<=fi<=b.

样例输入:

10 100
1234567890 9876543210
0 0

样例输出:

5
4

解题代码:

import java.io.BufferedInputStream;   
import java.math.BigInteger;   
import java.util.Scanner;   
  
/**  
 *  
 * poj2413  
 * @author NC  
 */  
public class Main {   
  
    public static void main(String[] args) {   
        Scanner scan = new Scanner(new BufferedInputStream(System.in));   
        while (scan.hasNext()) {   
            BigInteger a = scan.nextBigInteger();   
            BigInteger b = scan.nextBigInteger();   
            if (a.compareTo(BigInteger.ONE) == -1 && b.compareTo(BigInteger.ONE) == -1) {   
                break;   
            }   
            int count = 0;   
            BigInteger[] f = new BigInteger[1000000];   
            f[0] = BigInteger.ONE;   
            f[1] = BigInteger.ONE.add(BigInteger.ONE);   
            for (int i = 0; i < f.length; i++) {   
                if (i >= 2) {   
                    f[i] = f[i - 1].add(f[i - 2]);   
                }   
                if (f[i].compareTo(b)==1) {   
                    break;   
                }   
                if (!(f[i].compareTo(a)==-1)) {   
                    count++;   
                }   
            }   
            System.out.println(count);   
        }   
    }   
}

  1. 第一题是不是可以这样想,生了n孩子的家庭等价于n个家庭各生了一个1个孩子,这样最后男女的比例还是1:1

  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?