小牛生产小牛的问题解决集粹
问题:
一只刚出生的小牛,4年后生一只小牛,以后每年生一只。现有一只刚出生的小牛,问N年后共有牛多少只?
1.原始笨方法
private int Comput(int years)
{
//初始化为1头牛
int count = 1;
if (years < 4)
{
return count;
}
while (years - 3 > 0)
{
count = count + Comput(years - 3);
years--;
}
return count;
![](/Images/OutliningIndicators/InBlock.gif)
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}
特点:容易理解,不过效率太低.不具有实用价值.
2.采用HashTable优化方案
Hashtable table = new Hashtable();
public long Compute(int years)
{
//初始化为1头牛
long count = 1;
if (years <= 3)
{
return count;
}
int i = 4;
while (i <= years)
{
int subYears = i - 3;
if (table.ContainsKey(subYears))
{
count = (long)table[subYears];
}
else
{
count += Compute((int)(subYears));
}
if (!table.ContainsKey(subYears))
{
table.Add(subYears, count);
}
i++;
}
return (long)count;
}
特点:在第一种方案的基础上性能有了大幅度的提高.采用HashTable存储老牛的生育曲线,从而达到以后的小牛重复利用老牛的生育曲线.(直接取其生产数量)
3.采用数组的方式描述
public int Compute(int years)
{
int[] age = new int[4] { 1, 0, 0, 0 };
for (int i = 2; i <= years; i++)
{
age[3] += age[2]; age[2] = age[1];
age[1] = age[0];
age[0] = age[3];
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}
return (age[0] + age[1] + age[2] + age[3]);
}
特点:只采用一个循环搞定,效率极高.
3.采用优化递归公式实现
f(n) = f(n-1)+f(n-3) [n > 3]
f(n) = 1 [0 < n <= 3]
public int Comput(int x)
{
if (x < 4) return 1;
else return Comput(x - 1) + Comput(x - 3);
}
特点:代码简洁,功能简单实现,但使用递归当然会牺牲一定的效率作为代价.
前些天在网络上偶然发现的生产小牛问题.于是搜集整理了一下,方便大家共同研究学习使用.
有好的算法大家共同研究^_^
一只刚出生的小牛,4年后生一只小牛,以后每年生一只。现有一只刚出生的小牛,问N年后共有牛多少只?
1.原始笨方法
![](/Images/OutliningIndicators/None.gif)
![](/Images/OutliningIndicators/ExpandedBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
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![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedBlockEnd.gif)
2.采用HashTable优化方案
![](/Images/OutliningIndicators/None.gif)
![](/Images/OutliningIndicators/None.gif)
![](/Images/OutliningIndicators/ExpandedBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
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![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedBlockEnd.gif)
3.采用数组的方式描述
![](/Images/OutliningIndicators/None.gif)
![](/Images/OutliningIndicators/ExpandedBlockStart.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedSubBlockEnd.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedBlockEnd.gif)
特点:只采用一个循环搞定,效率极高.
3.采用优化递归公式实现
f(n) = f(n-1)+f(n-3) [n > 3]
f(n) = 1 [0 < n <= 3]
![](/Images/OutliningIndicators/None.gif)
![](/Images/OutliningIndicators/ExpandedBlockStart.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/InBlock.gif)
![](/Images/OutliningIndicators/ExpandedBlockEnd.gif)
特点:代码简洁,功能简单实现,但使用递归当然会牺牲一定的效率作为代价.
前些天在网络上偶然发现的生产小牛问题.于是搜集整理了一下,方便大家共同研究学习使用.
有好的算法大家共同研究^_^