Java 中的煎饼数

13 May 2025 | 5 分钟阅读

在本节中，我们将讨论什么是煎饼数，并使用不同的方法创建Java 程序来查找煎饼数。煎饼数程序经常在 Java 编码面试和学术中出现。

煎饼数

煎饼数 P_j 表示使用 j 次切割可以将圆形或煎饼分割成的最大块数。煎饼数也称为懒人割饼数，更正式的名称是中心多边形数。

在数学上，煎饼数 P_j 表示为

下图显示了 j = 0 到 j = 5 的煎饼数。

煎饼数 Java 程序

以下代码显示了使用上述数学公式实现的煎饼数。

文件名： PancakeNumberExample.java

public class PancakeNumberExample  
{

// main method 
public static void main (String argvs[])
{
int n = 10;

// loop for finding and printing
// the first 10 pancake numbers
for(int j = 0; j < n; j++)
{
// using the mathematical formula
int p = ((j * j) + j + 2) / 2;
System.out.print("For j = " + j);
System.out.println(", the pancake number is: " + p);
}

}
}

输出

For j = 0, the pancake number is: 1
For j = 1, the pancake number is: 2
For j = 2, the pancake number is: 4
For j = 3, the pancake number is: 7
For j = 4, the pancake number is: 11
For j = 5, the pancake number is: 16
For j = 6, the pancake number is: 22
For j = 7, the pancake number is: 29
For j = 8, the pancake number is: 37
For j = 9, the pancake number is: 46

使用组合计算煎饼数

煎饼数也可以通过组合来获得。使用组合，煎饼数定义为

P_j = ⁰C₀ = 1，对于 j = 0

P_j = ¹C₀ + ¹C₁ = 1 + 1 = 2，对于 j = 1

对于 j >= 2，

P_j = ^jC₀ + ^jC₁ + ^jC₂

因此，

对于 j = 2，

P₂ = ²C₀ + ²C₁ + ²C₂ = 1 + 2 + 1 = 4

对于 j = 3，

P₃ = ³C₀ + ³C₁ + ³C₂ = 1 + 3 + 3 = 7

对于 j = 4，

P₄ = ⁴C₀ + ⁴C₁ + ⁴C₂ = 1 + 4 + 6 = 11

对于 j = 5，

P₅ = ⁵C₀ + ⁵C₁ + ⁵C₂ = 1 + 5 + 10 = 16

将上述公式合并，对于 j = 0, 1, 和 >= 2，我们得到

P_j = 1，对于 j = 0，以及

P_j = ^{j + 1}C₂ + 1，对于 j >= 1

因此，

P₀ = 1

P₁ = ^{1 + 1}C₂ + 1 = ²C₂ + 1 = 1 + 1 = 2

P₂ = ^{2 + 1}C₂ + 1 = ³C₂ + 1 = 3 + 1 = 4

P₃ = ^{3 + 1}C₂ + 1 = ⁴C₂ + 1 = 6 + 1 = 7

煎饼数：迭代方法

让我们来实现上述组合公式来计算煎饼数。

文件名： PancakeNumberExample1.java

public class PancakeNumberExample1  
{
public int findPancakeNo(int j)
{
// handling the base case
if(j == 0)
{
return 1;
}

int num = 1;

// for calculating (j + 1)!
for(int c = 1; c <= j + 1; c++)
{
num = num * c;
}

int denom = 1;

// for calculating ((j + 1) - 2)!
for(int c = 1; c <= j - 1; c++)
{
denom = denom * c;
}

// caculating (j + 1)!/(((j + 1) - 2)! * 2!) + 1
num = num / denom;
num = num / 2;
num = num + 1;


// returning the computed pancake number
return num;


}

// main method 
public static void main (String argvs[])
{
int n = 10;

// creating an object of the class PancakeNumberExample1
PancakeNumberExample1 obj = new PancakeNumberExample1();


// loop for finding and printing
// the first 10 pancake numbers
for(int j = 0; j < n; j++)
{
// invoking the method findPancakeNo()
int p = obj.findPancakeNo(j);
System.out.print("For j = " + j);
System.out.println(", the pancake number is: " + p);
}

}
}

输出

For j = 0, the pancake number is: 1
For j = 1, the pancake number is: 2
For j = 2, the pancake number is: 4
For j = 3, the pancake number is: 7
For j = 4, the pancake number is: 11
For j = 5, the pancake number is: 16
For j = 6, the pancake number is: 22
For j = 7, the pancake number is: 29
For j = 8, the pancake number is: 37
For j = 9, the pancake number is: 46

煎饼数：递归方法

组合公式也可以通过递归来实现。下面的程序显示了这一点。

文件名： PancakeNumberExample2.java

public class PancakeNumberExample2  
{
public int findPancakeNo(int j, int b)
{
// handling the base case
if(j == 0 || b == 0)
{
return 1;
}


// The recursion is based on the nCr = n / r * (n - 1)C(r - 1)
// property of combination
int temp = findPancakeNo(j - 1, b - 1);
temp = (temp * j) / b;
return temp;

}

// main method 
public static void main (String argvs[])
{
int n = 10;

// creating an object of the class PancakeNumberExample2
PancakeNumberExample2 obj = new PancakeNumberExample2();


// loop for finding and printing
// the first 10 pancake numbers
for(int j = 0; j < n; j++)
{
// invoking the method findPancakeNo()
int p;

if(j == 0)
{
// for j = 0, p = 1
p = 1;
}
else
{
p = obj.findPancakeNo(j + 1, 2) + 1;
}

System.out.print("For j = " + j);
System.out.println(", the pancake number is: " + p);
}

}
}

输出

For j = 0, the pancake number is: 1
For j = 1, the pancake number is: 2
For j = 2, the pancake number is: 4
For j = 3, the pancake number is: 7
For j = 4, the pancake number is: 11
For j = 5, the pancake number is: 16
For j = 6, the pancake number is: 22
For j = 7, the pancake number is: 29
For j = 8, the pancake number is: 37
For j = 9, the pancake number is: 46

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Java 中的煎饼数

煎饼数

煎饼数 Java 程序

使用组合计算煎饼数

煎饼数：迭代方法

煎饼数：递归方法

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Java 中的煎饼数

煎饼数

煎饼数 Java 程序

使用组合计算煎饼数

煎饼数：迭代方法

煎饼数：递归方法

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