标签:效果 param water order 斐波那契数 baidu nts 引入 row
斐波那契数列指的是这样一个数列 0, 1, 1, 2,3, 5, 8, 13, 21, 34, 55, 89, 144, ...对于这个数列仅仅能说将兔子生产周期第为3月。假设生成周期变成4月这个数列肯定不是这种,或者说兔子还有死亡周期,在这里我是对兔子生产周期没有限定。仅仅要月份大于生产周期都能够计算出第month月份究竟能产生多少对兔子。
Java兔子生殖问题
斐波那契数列又因数学家列昂纳多·斐波那契以兔子生殖为样例而引入。故又称为“兔子数列”。
一般而言,兔子在出生两个月后。就有生殖能力。一对兔子每一个月能生出一对小兔子来。假设全部兔子都不死,那么一年以后能够生殖多少对兔子?
我们最好还是拿新出生的一对小兔子分析一下:
第一个月小兔子没有生殖能力,所以还是一对
两个月后,生下一对小兔对数共同拥有两对
三个月以后。老兔子又生下一对。由于小兔子还没有生殖能力。所以一共是三对。
------
依次类推能够列出下表:
经过月数 |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
幼仔对数 |
1 |
0 |
1 |
1 |
2 |
3 |
5 |
8 |
13 |
21 |
34 |
55 |
89 |
成兔对数 |
0 |
1 |
1 |
2 |
3 |
5 |
8 |
13 |
21 |
34 |
55 |
89 |
144 |
整体对数 |
1 |
1 |
2 |
3 |
5 |
8 |
13 |
21 |
34 |
55 |
89 |
144 |
233 |
幼仔对数=前月成兔对数
成兔对数=前月成兔对数+前月幼仔对数
整体对数=本月成兔对数+本月幼仔对数
能够看出幼仔对数、成兔对数、整体对数都构成了一个数列。这个数列有关十分明显的特点,那是:前面相邻两项之和,构成了后一项。
这个数列是意大利中世纪数学家斐波那契在<算盘全书>中提出的。这个级数的通项公式,除了具有a(n+2)=an+a(n+1)的性质外,还能够证明通项公式为:an=(1/√5)*{[(1+√5)/2]^n-[(1-√5)/2]^n}(n=1,2,3.....)。
JavaCode:
/* * System Abbrev : * system Name : * Component No : * Component Name: * File name :FabonacciSequence.java * Author :Peter.Qiu * Date :Aug 25, 2014 * Description : <description> */
/* Updation record 1: * Updation date : Aug 25, 2014 * Updator : Peter.Qiu * Trace No: <Trace No> * Updation No: <Updation No> * Updation Content: <List all contents of updation and all methods updated.> */ package com.qiuzhping.util.interesting;
/** * <Description functions in a word> * <Detail description> * * @author Peter.Qiu * @version [Version NO, Aug 25, 2014] * @see [Related classes/methods] * @since [product/module version] */ public class FabonacciSequence {
private static FabonacciSequence util = null;
/** <default constructor> */ public FabonacciSequence() { // TODO Auto-generated constructor stub }
/** <Description functions in a word> * Aug 25, 2014 * <Detail description> * @author Peter.Qiu * @param args [Parameters description] * @return void [Return type description] * @exception throws [Exception] [Exception description] * @see [Related classes#Related methods#Related properties] */ public static void main(String[] args) { long month = 8; long product = 4; long start = System.currentTimeMillis(); // System.out.println(" fabonacci \trabbit : "+getInstance().fabonacci(month)); // System.out.println(" take times = "+(System.currentTimeMillis() - start)/1000); // System.out.println("--------------------------------------"); // System.out.println(" fabonacci1 \trabbit : "+getInstance().fabonacci1(month)); // System.out.println(" take times = "+(System.currentTimeMillis() - start)/1000); // System.out.println("--------------------------------------"); // System.out.println(" fabonacci2 \trabbit : "+getInstance().fabonacci2(month,product)); // System.out.println(" take times = "+(System.currentTimeMillis() - start)/1000);
for(long i = product; i <= month; i++){ System.out.println("month = "+i+"\tfabonacci2 \trabbit : "+getInstance().fabonacci2(i,product)); } }
public static FabonacciSequence getInstance() { if (util == null) { util = new FabonacciSequence(); } return util; }
/** <Description functions in a word> *pruduct month = 3<BR> * Aug 25, 2014 * <Detail description> * @author Peter.Qiu * @param month : How many months. * @return [Parameters description] * @return long [Return type description] * @exception throws [Exception] [Exception description] * @see [Related classes#Related methods#Related properties] */ public long fabonacci(long month) { if (!(month < 3)) { for (long i = 3; i <= month;) { return fabonacci(month - 1) + fabonacci(month - 2); } } return 1; }
/** <Description functions in a word> * pruduct month = 3<BR> * Aug 25, 2014 * <Detail description> * @author Peter.Qiu * @param month :How many months. * @return [Parameters description] * @return long [Return type description] * @exception throws [Exception] [Exception description] * @see [Related classes#Related methods#Related properties] */ public long fabonacci1(long month) { long sum = 1, lastMonth = 1, lastLastMonth = 1; if (!(month < 3)) { for (long i = 3; i <= month; i++) { lastLastMonth = lastMonth; lastMonth = sum; sum = lastLastMonth + lastMonth; } } return sum; }
/** <Description functions in a word> * Aug 25, 2014 * <Detail description> * @author Peter.Qiu * @param month:How many months. * @param pruductMonth:The production cycle. * @return [Parameters description] * @return long [Return type description] * @exception throws [Exception] [Exception description] * @see [Related classes#Related methods#Related properties] */ public long fabonacci2(long month, long pruductMonth) { long sum = 1; if (!(month < pruductMonth)) { for (long i = 0,j = pruductMonth - 1; i < month - (pruductMonth - 1); i++) { sum += fabonacci2(month - j - i, pruductMonth); } } return sum; } } |
以上这些算法都没有应用到面向对象的思维方式去解决,假设生成周期变成4月这个数列肯定不是这种,或者说兔子还有死亡周期,在这里我是对兔子生产周期没有限定,仅仅要月份大于生产周期都能够计算出第month月份究竟能产生多少对兔子。实际中能够将兔子定于为一个Rabbit对象,将其生产周期、死亡周期定义为属性。可能这个能有更好的效果。
标签:效果 param water order 斐波那契数 baidu nts 引入 row
原文地址:http://www.cnblogs.com/llguanli/p/6869588.html