给定的整数N个台阶的，任务是计数数量的方式通过取1或2步骤任意次数，但服用3步骤一次以到达第N楼梯。
例子：
Input: N = 4
Output: 2
Explanation:
Since a step of 3 has to be taken compulsorily and only once. So, there are only two possible ways: (1, 3) or (3, 1)
Input: N = 5
Output: 5
Explanation:
There are five possible ways to reach 5th stair: (2, 3), (3, 2), (1, 1, 3), (1, 3, 1), (3, 1, 1)
方法：可以使用置换和组合的概念来解决此问题。存在一个约束，根据该约束，一次必须精确执行一次三个步骤。这个想法是要计算可能要采取的3个步骤的职位数量。
现在，下一个任务是找到移动到第N个楼梯的可能的两个步骤的数量，即(N – 3)/2。这样，如果计数为两个，则所有可能的方法都将被覆盖。步骤一次增加一次，并且对于每个步骤，将计算所有可能的组合。
在概括了步骤之后，每个可能序列的可能组合将是
Ways = (Length of sequence)! * (Count of 2-step)!
------------------------------------------
(Count of 1-step)! 算法：
- 将序列的长度初始化为(N – 2)。
- 将1步的计数初始化为(N – 3)。
- 将2步计数初始化为0
- 将可能的2步计数初始化为(N-3)/ 2。
- 循环运行，直到可能的2步计数等于2步的实际计数。
- 借助当前序列的长度，1步计数，2步计数以及上述公式，找到可能的方法数量。
- 将序列的长度减少1。
- 将2步计数增加1
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 or 2 steps at a time and
// 3rd step exactly once
#include
using namespace std;
int factorial(int n)
{
// Single line to find factorial
return (n == 1 || n == 0) ? 1 :
n * factorial(n - 1);
}
// Function to find
// the number of ways
int ways(int n)
{
// Base Case
if (n < 3)
{
return 0;
}
// Count of 2-steps
int c2 = 0;
// Count of 1-steps
int c1 = n - 3;
// Intial length of sequence
int l = c1 + 1;
int s = 0;
// Expected count of 2-steps
int exp_c2 = c1 / 2;
// Loop to find the ways for
// every possible sequence
while (exp_c2 >= c2)
{
int f1 = factorial(l);
int f2 = factorial(c1);
int f3 = factorial(c2);
int f4 = (f2 * f3);
s += f1 / f4;
c2 += 1;
c1 -= 2;
l -= 1;
}
return s;
}
// Driver code
int main()
{
int n = 7;
int ans = ways(n);
cout << ans << endl;
return 0;
}
// This code is contributed by coder001 Java
// Java implementation to find the number
// the number of ways to reach Nth stair
// by taking 1 or 2 steps at a time and
// 3rd step exactly once
class GFG {
static int factorial(int n)
{
// Single line to find factorial
return (n == 1 || n == 0) ? 1 :
n * factorial(n - 1);
}
// Function to find
// the number of ways
static int ways(int n)
{
// Base Case
if (n < 3)
{
return 0;
}
// Count of 2-steps
int c2 = 0;
// Count of 1-steps
int c1 = n - 3;
// Intial length of sequence
int l = c1 + 1;
int s = 0;
// Expected count of 2-steps
int exp_c2 = c1 / 2;
// Loop to find the ways for
// every possible sequence
while (exp_c2 >= c2)
{
int f1 = factorial(l);
int f2 = factorial(c1);
int f3 = factorial(c2);
int f4 = (f2 * f3);
s += f1 / f4;
c2 += 1;
c1 -= 2;
l -= 1;
}
return s;
}
// Driver Code
public static void main(String args[])
{
int n = 7;
int ans = ways(n);
System.out.println(ans);
}
}
// This code is contributed by rutvik_56Python3
# Python3 implementation to find the number
# the number of ways to reach Nth stair
# by taking 1 or 2 steps at a time and
# 3rd Step exactly once
import math
# Function to find
# the number of ways
def ways(n):
# Base Case
if n < 3:
return 0
# Count of 2-steps
c2 = 0
# Count of 1-steps
c1 = n-3
# Intial length of sequence
l = c1 + 1
s = 0
# expected count of 2-steps
exp_c2 = c1 / 2
# Loop to find the ways for
# every possible sequence
while exp_c2 >= c2:
f1 = math.factorial(l)
f2 = math.factorial(c1)
f3 = math.factorial(c2)
s += f1//(f2 * f3)
c2 += 1
c1 -= 2
l -= 1
return s
# Driver Code
if __name__ == "__main__":
N = 7
print(ways(N))C#
// C# implementation to find the number
// the number of ways to reach Nth stair
// by taking 1 or 2 steps at a time and
// 3rd step exactly once
using System;
class GFG{
static int factorial(int n)
{
// Single line to find factorial
return (n == 1 || n == 0) ? 1 :
n * factorial(n - 1);
}
// Function to find
// the number of ways
static int ways(int n)
{
// Base Case
if (n < 3)
{
return 0;
}
// Count of 2-steps
int c2 = 0;
// Count of 1-steps
int c1 = n - 3;
// Intial length of sequence
int l = c1 + 1;
int s = 0;
// Expected count of 2-steps
int exp_c2 = c1 / 2;
// Loop to find the ways for
// every possible sequence
while (exp_c2 >= c2)
{
int f1 = factorial(l);
int f2 = factorial(c1);
int f3 = factorial(c2);
int f4 = (f2 * f3);
s += f1 / f4;
c2 += 1;
c1 -= 2;
l -= 1;
}
return s;
}
// Driver code
static void Main()
{
int n = 7;
int ans = ways(n);
Console.WriteLine(ans);
}
}
// This code is contributed by divyeshrabadiya07Javascript
输出：
20性能分析：
- 时间复杂度：与上述方法一样，有一个循环来检查每个序列的可能排列，直到可能花费O(N)的上至(N-3)为止，并找出可能要花费O(N)的组合，因此时间复杂度将为O(N 2 ) 。
- 空间复杂度：与上述方法一样，没有使用额外的空间，因此空间复杂度将为O(1) 。
注意：通过预先计算直到N的每个数字的阶乘，可以将时间复杂度提高到O(N)。
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