计算通过 1 步和 2 步到达第 N 级楼梯的方法，正好是 3 步

给定一个表示楼梯数量的数字N ，任务是通过走 1、2 步任意次数到达第N^个楼梯，并且只走 3 步一次。
例子：

Input: N = 4 
Output: 2 
Explanation: 
Since a step of 3 has to be taken compulsorily and only once, 
There are only two possible ways: (1, 3) or (3, 1)
Input: N = 5 
Output: 5 
Explanation: 
Since a step of 3 has to be taken compulsorily and only once, 
There are only 5 possible ways: 
(1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 3), (3, 2) 
 

方法：这个问题可以用动态规划解决。要到达^第N^个楼梯，该人应位于 (N – 1) ^th 、(N – 2) ^th或 (N – 3) ^th 。所以，为了达到n的底座延伸(N – 1)^个楼梯^个楼梯，其包括精确只的3一个步骤，前移(N – 2)与步骤^个步骤，其中包括的3正好一个步骤并到达(N – 3)^第3 步，不采取任何 3 步。
所以，这个问题的递归关系将是——

includes_3[i] = includes_3[i-1] + includes_3[i-2] + not_includes[i-3] 
 

而不允许一次执行 3 个步骤时的递归关系为

not_includes[i] = not_includes[i – 1] + not_includes[i – 2] 
 

下面是上述方法的实现。

C++

// C++ implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
 
#include 
using namespace std;
 
// Function to find the number
// the number of ways to reach Nth stair
int number_of_ways(int n)
{
    // Array including number
    // of ways that includes 3
    int includes_3[n + 1] = {};
 
    // Array including number of
    // ways that doesn't includes 3
    int not_includes_3[n + 1] = {};
 
    // Initially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (int i = 4; i <= n; i++) {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] + not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
int main()
{
    int n = 7;
 
    cout << number_of_ways(n);
 
    return 0;
}

Java

// Java implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
class GFG
{
 
// Function to find the number
// the number of ways to reach Nth stair
static int number_of_ways(int n)
{
    // Array including number
    // of ways that includes 3
    int []includes_3 = new int[n + 1];
     
    // Array including number of
    // ways that doesn't includes 3
    int []not_includes_3 = new int[n + 1];
 
    // Initially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (int i = 4; i <= n; i++)
    {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] +
               not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 7;
 
    System.out.print(number_of_ways(n));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to find the number
# the number of ways to reach Nth stair
# by taking 1, 2 step at a time and
# 3 Steps at a time exactly once.
 
# Function to find the number
# the number of ways to reach Nth stair
def number_of_ways(n):
     
    # Array including number
    # of ways that includes 3
    includes_3 = [0]*(n + 1)
 
    # Array including number of
    # ways that doesn't includes 3
    not_includes_3 = [0] * (n + 1)
 
    # Initially to reach 3 stairs by
    # taking 3 steps can be
    # reached by 1 way
    includes_3[3] = 1
 
    not_includes_3[1] = 1
    not_includes_3[2] = 2
    not_includes_3[3] = 3
 
    # Loop to find the number
    # the number of ways to reach Nth stair
    for i in range(4, n + 1):
        includes_3[i] = includes_3[i - 1] + \
                        includes_3[i - 2] + \
                        not_includes_3[i - 3]
        not_includes_3[i] = not_includes_3[i - 1] + \
                           not_includes_3[i - 2]
    return includes_3[n]
 
# Driver Code
n = 7
 
print(number_of_ways(n))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
using System;
 
class GFG
{
 
// Function to find the number
// the number of ways to reach Nth stair
static int number_of_ways(int n)
{
    // Array including number
    // of ways that includes 3
    int []includes_3 = new int[n + 1];
     
    // Array including number of
    // ways that doesn't includes 3
    int []not_includes_3 = new int[n + 1];
 
    // Initially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (int i = 4; i <= n; i++)
    {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] +
            not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 7;
 
    Console.Write(number_of_ways(n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
 
// Function to find the number
// the number of ways to reach Nth stair
function number_of_ways(n)
{
    // Array including number
    // of ways that includes 3
    let includes_3 = new Uint8Array(n + 1);
 
    // Array including number of
    // ways that doesn't includes 3
    let not_includes_3 = new Uint8Array(n + 1);
 
    // Initially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (let i = 4; i <= n; i++) {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] + not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
 
    let n = 7;
 
    document.write(number_of_ways(n));
 
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

20

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