爬上第n个楼梯，允许从1到n的所有跳跃(三种不同的方法)

一只猴子正站在下面有N个台阶的楼梯上。考虑到一次可能需要1到N步的跳跃，请计算有多少种方法可以到达楼梯的顶部?

例子：

Input : 2
Output : 2
It can either take (1, 1) or (2) to
reach the top. So, total 2 ways

Input : 3
Output : 4
Possibilities : (1, 1, 1), (1, 2), (2, 1),
(3). So, total 4 ways

有3种不同的方式来考虑问题。

1. 在所有可能的解决方案中，猴子要么踩到一步，要么可以跳过。因此，使用基本计数原理，第一步有2种参与方式，对于每种方式，第二步也有2种参与方式，依此类推。但是最后一步总是必须踩的。

   2 x 2 x 2 x .... x 2(N-1 th step) x 1(Nth step) 
     = 2(N-1) different ways. 
   

2. 让我们为用例定义一个函数F(n)。 F(n)表示从底部到顶部有N个台阶的所有可能的到达方式，其中最小跃点为1步，最大跃点为N步。现在，对于猴子来说，它可以进行的第一个举动有N种不同的方式(1步，2步，3步…. N步)。如果将第一个跃点作为第一步，它将剩下N-1个要征服的步骤，可以通过F(N-1)方式实现。如果将第一个飞跃作为2步，它将覆盖N-2步，这可以用F(N-2)方式实现。放在一起，

   F(N) = F(N-1) + F(N-2) + F(N-3) + ... + 
                         F(2) + F(1) + F(0) 
   Now, 
   F(0) = 1
   F(1) = 1
   F(2) = 2
   F(3) = 4
   
   Hence,
   F(N) = 1 + 1 + 2 + 4 + ... + F(n-1)
        = 1 + 2^0 + 2^1 + 2^2 + ... + 2^(n-2)
        = 1 + [2^(n-1) - 1]
   

   C++

   // C++ program to count total number of ways
   // to reach n-th stair with all jumps alowed
   #include 
     
   int calculateLeaps(int n)
   {
       if (n == 0 || n == 1) {
           return 1;
       }
       else {
           int leaps = 0;
           for (int i = 0; i < n; i++)
               leaps += calculateLeaps(i);
           return leaps;
       }
   }
     
   // Driver code
   int main()
   {
       int calculateLeaps(int);
       std::cout << calculateLeaps(4) << std::endl;
       return 0;
   }

   Java

   // Java program to count total number of ways
   // to reach n-th stair with all jumps alowed
   class GFG {
       static int calculateLeaps(int n)
       {
           if (n == 0 || n == 1) {
               return 1;
           }
           else {
               int leaps = 0;
               for (int i = 0; i < n; i++)
                   leaps += calculateLeaps(i);
               return leaps;
           }
       }
     
       // Driver code
       public static void main(String[] args)
       {
           System.out.println(calculateLeaps(4));
       }
   }
   // This code is contributed by Anant Agarwal.

   Python3

   # Python program to count
   # total number of ways
   # to reach n-th stair with
   # all jumps alowed
     
   def calculateLeaps(n):
       if n == 0 or n == 1:
           return 1;
       else:
           leaps = 0;
           for i in range(0,n):
               leaps = leaps + calculateLeaps(i);
           return leaps;
     
   # Driver code
   print(calculateLeaps(4));
     
   # This code is contributed by mits

   C#

   // C# program to count total number of ways
   // to reach n-th stair with all jumps alowed
   using System;
     
   class GFG {
     
       // Function to calculate leaps
       static int calculateLeaps(int n)
       {
           if (n == 0 || n == 1) {
               return 1;
           }
           else {
               int leaps = 0;
               for (int i = 0; i < n; i++)
                   leaps += calculateLeaps(i);
               return leaps;
           }
       }
     
       // Driver code
       public static void Main()
       {
           Console.WriteLine(calculateLeaps(4));
       }
   }
     
   // This code is contributed by vt_m.

   输出：

   8
   

   通过使用动态编程可以改善上述解决方案

3. 让我们将此问题分解为一些小问题。猴子必须踩到最后一步，前N-1个步骤是可选的。猴子可以先踩0步，然后再踩到最高步，这是最大的跳跃。或者，它可以决定仅在两者之间踩一次，这可以通过n-1种方式[ ^(N-1) C ₁ ]实现。依此类推，在以^(N-1) C ₂方式到达顶部之前，它只能踩2步。放在一起..

   F(N)= ^(N-1) C ₀ + ^(N-1) C ₁ + ^(N-1) C ₂ + … + ^(N-1) C _(N-2) + ^(N-1) C _(N-1)
   这是二项式系数之和。
   = 2 ^(n-1)

C++

// C++ program to count total number of ways
// to reach n-th stair with all jumps alowed
#include 
  
int calculateLeaps(int n)
{
    if (n == 0)
        return 1;
    return (1 << (n - 1));
}
  
// Driver code
int main()
{
    int calculateLeaps(int);
    std::cout << calculateLeaps(4) << std::endl;
    return 0;
}

Java

// Java program to count total number of ways
// to reach n-th stair with all jumps alowed
class GFG {
    static int calculateLeaps(int n)
    {
        if (n == 0)
            return 1;
        return (1 << (n - 1));
    }
  
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(calculateLeaps(4));
    }
}
// This code is contributed by Anant Agarwal.

Python3

# python3 program to count
# total number of ways
# to reach n-th stair with
# all jumps alowed
  
def calculateLeaps(n):
    if (n == 0):
        return 1;
    return (1 << (n - 1));
  
# Driver code
print(calculateLeaps(4));
  
# This code is contributed
# by mits

C＃

// C# program to count total number of ways
// to reach n-th stair with all jumps alowed
using System;
  
class GFG {
  
    // Function to calculate leaps
    static int calculateLeaps(int n)
    {
        if (n == 0)
            return 1;
        return (1 << (n - 1));
    }
  
    // Driver code
    public static void Main()
    {
        Console.WriteLine(calculateLeaps(4));
    }
}
  
// This code is contributed by vt_m.

输出：

8