计算从起始石到达终点的方式，每一步最多跳 K 次

从左到右连续给定N块石头。从每块石头，你最多可以跳到K块石头。任务是找到办法从某物石到第N石达到总数。
例子：

Input: N = 5, s = 2, K = 2 
Output: Total Ways = 3 
Explanation: 
Assume s1, s2, s3, s4, s5 be the stones. The possible paths from 2nd stone to 5th stone: 
s2 -> s3 -> s4 -> s5 
s2 -> s4 -> s5 
s2 -> s3 -> s5 
Hence total number of ways = 3
Input: N = 8, s = 1, K = 3 
Output: Total Ways = 44 
 

方法：

1. 假设dp[i]是到达第 i 个石头的方法数。
2. 因为最多有K 次跳跃，所以第i 个石头可以被它之前的所有K个石头到达。
3. 迭代所有可能的K 次跳跃，并继续将此可能的组合添加到数组dp[] 。
4. 那么从第s块石头到达第 N 个节点的可能方法总数将是dp[N-1] 。
5. 例如：

Let N = 5, s = 2, K = 2, then we have to reach Nth stone from sth stone. 
Let dp[N+1] is the array that stores the number of paths to reach the Nth Node from sth stone. 
Initially, dp[] = { 0, 0, 0, 0, 0, 0 } and dp[s] = 1, then 
dp[] = { 0, 0, 1, 0, 0, 0 } 
To reach the 3rd, 
There is only 1 way with at most 2 jumps i.e., from stone 2(with jump = 1). Update dp[3] = dp[2] 
dp[] = { 0, 0, 1, 1, 0, 0 }
To reach the 4th stone, 
The two ways with at most 2 jumps i.e., from stone 2(with jump = 2) and stone 3(jump = 1). Update dp[4] = dp[3] + dp[2] 
dp[] = { 0, 0, 1, 1, 2, 0 }
To reach the 5th stone, 
The two ways with at most 2 jumps i.e., from stone 3(with jump = 2) and stone 4(with jump = 1). Update dp[5] = dp[4] + dp[3] 
dp[] = { 0, 0, 1, 1, 2, 3 }
Now dp[N] = 3 is the number of ways to reach Nth stone from sth stone. 
 

下面是上述方法的实现：

C++

// C++ program to find total no.of ways
// to reach nth step
#include "bits/stdc++.h"
using namespace std;
 
// Function which returns total no.of ways
// to reach nth step from sth steps
int TotalWays(int n, int s, int k)
{
    // Initialize dp array
    int dp[n];
 
    // filling all the elements with 0
    memset(dp, 0, sizeof(dp));
 
    // Initialize (s-1)th index to 1
    dp[s - 1] = 1;
 
    // Iterate a loop from s to n
    for (int i = s; i < n; i++) {
 
        // starting range for counting ranges
        int idx = max(s - 1, i - k);
 
        // Calculate Maximum moves to
        // Reach ith step
        for (int j = idx; j < i; j++) {
            dp[i] += dp[j];
        }
    }
 
    // For nth step return dp[n-1]
    return dp[n - 1];
}
 
// Driver Code
int main()
{
    // no of steps
    int n = 5;
 
    // Atmost steps allowed
    int k = 2;
 
    // starting range
    int s = 2;
    cout << "Total Ways = "
         << TotalWays(n, s, k);
}

Java

// Java program to find total no.of ways
// to reach nth step
class GFG{
  
// Function which returns total no.of ways
// to reach nth step from sth steps
static int TotalWays(int n, int s, int k)
{
    // Initialize dp array
    int []dp = new int[n];
  
    // Initialize (s-1)th index to 1
    dp[s - 1] = 1;
  
    // Iterate a loop from s to n
    for (int i = s; i < n; i++) {
  
        // starting range for counting ranges
        int idx = Math.max(s - 1, i - k);
  
        // Calculate Maximum moves to
        // Reach ith step
        for (int j = idx; j < i; j++) {
            dp[i] += dp[j];
        }
    }
  
    // For nth step return dp[n-1]
    return dp[n - 1];
}
  
// Driver Code
public static void main(String[] args)
{
    // no of steps
    int n = 5;
  
    // Atmost steps allowed
    int k = 2;
  
    // starting range
    int s = 2;
    System.out.print("Total Ways = "
         + TotalWays(n, s, k));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python 3 program to find total no.of ways
# to reach nth step
 
# Function which returns total no.of ways
# to reach nth step from sth steps
def TotalWays(n, s, k):
 
    # Initialize dp array
    dp = [0]*n
 
    # Initialize (s-1)th index to 1
    dp[s - 1] = 1
 
    # Iterate a loop from s to n
    for i in range(s, n):
 
        # starting range for counting ranges
        idx = max(s - 1, i - k)
 
        # Calculate Maximum moves to
        # Reach ith step
        for j in range( idx, i) :
            dp[i] += dp[j]
 
    # For nth step return dp[n-1]
    return dp[n - 1]
 
# Driver Code
if __name__ == "__main__":
    # no of steps
    n = 5
 
    # Atmost steps allowed
    k = 2
 
    # starting range
    s = 2
    print("Total Ways = ", TotalWays(n, s, k))
     
# This code is contributed by chitranayal

C#

// C# program to find total no.of ways
// to reach nth step
using System;
 
class GFG{
      
    // Function which returns total no.of ways
    // to reach nth step from sth steps
    static int TotalWays(int n, int s, int k)
    {
        // Initialize dp array
        int []dp = new int[n];
      
        // Initialize (s-1)th index to 1
        dp[s - 1] = 1;
      
        // Iterate a loop from s to n
        for (int i = s; i < n; i++) {
      
            // starting range for counting ranges
            int idx = Math.Max(s - 1, i - k);
      
            // Calculate Maximum moves to
            // Reach ith step
            for (int j = idx; j < i; j++) {
                dp[i] += dp[j];
            }
        }
      
        // For nth step return dp[n-1]
        return dp[n - 1];
    }
      
    // Driver Code
    public static void Main(string[] args)
    {
        // no of steps
        int n = 5;
      
        // Atmost steps allowed
        int k = 2;
      
        // starting range
        int s = 2;
        Console.Write("Total Ways = "+ TotalWays(n, s, k));
    }
}
 
// This code is contributed by Yash_R

Javascript

// Javascript program to find total no.of ways
// to reach nth step
 
 
// Function which returns total no.of ways
// to reach nth step from sth steps
function TotalWays(n, s, k)
{
    // Initialize dp array
    let dp = new Array(n);
 
    // filling all the elements with 0
    dp.fill(0);
 
    // Initialize (s-1)th index to 1
    dp[s - 1] = 1;
 
    // Iterate a loop from s to n
    for (let i = s; i < n; i++) {
 
        // starting range for counting ranges
        let idx = Math.max(s - 1, i - k);
 
        // Calculate Maximum moves to
        // Reach ith step
        for (let j = idx; j < i; j++) {
            dp[i] += dp[j];
        }
    }
 
    // For nth step return dp[n-1]
    return dp[n - 1];
}
 
// Driver Code
 
 
    // no of steps
    let n = 5;
 
    // Atmost steps allowed
    let k = 2;
 
    // starting range
    let s = 2;
    document.write("Total Ways = " + TotalWays(n, s, k));
 
    // This code is contributed by _saurabh_jaiswal

输出：

Total Ways = 3

时间复杂度： O(N ² )，其中 N 是石头的数量。
辅助空间： O(N)

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