在 [1, N] 范围内长度为 K 的序列的计数，其中每个元素都是其前一个元素的倍数

给定两个整数N和K，任务是从范围[1, N]中找到K个元素的序列计数，其中每个元素都是前一个元素的倍数。

例子：

Input: N = 4, K = 3 
Output: 13
Explanation: The sequences that can be made from the integers 1, 2, 3, 4 having 3 elements are: {1, 1, 1}, {2, 2, 2}, {3, 3, 3}, {4, 4, 4}, {1, 1, 2}, {1, 2, 2}, {1, 2, 4}, {1, 1, 3}, {1, 3, 3}, {1, 1, 4}, {1, 4, 4}, {2, 2, 4}, and {2, 4, 4}.

Input: N = 9, K = 5 
Output: 111

方法：给定的问题可以使用带有记忆的递归来解决。请按照以下步骤解决问题：

• 创建一个 2D 数组dp[][]存储记忆状态，其中dp[i][j]表示长度为i的序列的计数，其中j作为第一个元素。
• 创建一个递归函数countSequenceUtil() ，它将序列的长度和起始元素作为参数，将下一个元素设置为当前元素的倍数，并递归调用剩余的序列。
• 将计算状态的答案存储在dp[][]数组中，如果某些状态的值已经计算，则返回它。
• 创建一个函数countSequence() ，它遍历序列中所有可能的起始元素，并调用递归函数来计算具有该起始元素的K个元素的序列。
• 保持变量ans中每个起始元素的计算计数总和，这是所需的值。

下面是上述方法的实现：

C++

// C++ implementation for the above approach
#include 
using namespace std;
 
// Initialize the dp matrix
int static dp[1001][1001];
 
// Function to find the count of sequences of K
// elements with first element as m where every
// element is a multiple of the previous one
int countSequenceUtil(int k, int m, int n)
{
    // Base case
    if (k == 1) {
        return 1;
    }
 
    // If the value already exists
    // in the DP then return it
    if (dp[k][m] != -1) {
        return dp[k][m];
    }
 
    // Variable to store the count
    int res = 0;
 
    for (int i = 1; i <= (n / m); i++) {
 
        // Recursive Call
        res += countSequenceUtil(k - 1,
                                 m * i, n);
    }
 
    // Store the calculated
    // answer and return it
    return dp[k][m] = res;
}
 
// Function to find count of sequences of K
// elements in the range [1, n] where every
// element is a multiple of the previous one
int countSequence(int N, int K)
{
    // Initializing all values
    // of dp with -1
    memset(dp, -1, sizeof(dp));
 
    // Variable to store
    // the total count
    int ans = 0;
 
    // Iterate from 1 to N
    for (int i = 1; i <= N; i++) {
        ans += countSequenceUtil(K, i, N);
    }
 
    // Return ans
    return ans;
}
 
// Driver Code
int main()
{
    int N = 9;
    int K = 5;
 
    cout << countSequence(N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
   
// Initialize the dp matrix
static int dp[][] = new int[1001][1001];
 
// Function to find the count of sequences of K
// elements with first element as m where every
// element is a multiple of the previous one
static int countSequenceUtil(int k, int m, int n)
{
    // Base case
    if (k == 1) {
        return 1;
    }
 
    // If the value already exists
    // in the DP then return it
    if (dp[k][m] != -1) {
        return dp[k][m];
    }
 
    // Variable to store the count
    int res = 0;
 
    for (int i = 1; i <= (n / m); i++) {
 
        // Recursive Call
        res += countSequenceUtil(k - 1,
                                 m * i, n);
    }
 
    // Store the calculated
    // answer and return it
    return dp[k][m] = res;
}
 
// Function to find count of sequences of K
// elements in the range [1, n] where every
// element is a multiple of the previous one
static int countSequence(int N, int K)
{
   
    // Initializing all values
    // of dp with -1
     for(int i=0;i

---

Python3

# Python implementation for the above approach
 
# Initialize the dp matrix
dp = [[-1 for i in range(1001)] for j in range(1001)]
 
# Function to find the count of sequences of K
# elements with first element as m where every
# element is a multiple of the previous one
def countSequenceUtil(k, m, n):
 
    # Base case
    if (k == 1):
        return 1
 
    # If the value already exists
    # in the DP then return it
    if (dp[k][m] != -1):
        return dp[k][m]
 
    # Variable to store the count
    res = 0
 
    for i in range(1, (n // m) + 1):
 
        # Recursive Call
        res += countSequenceUtil(k - 1,
                                 m * i, n)
 
    # Store the calculated
    # answer and return it
    dp[k][m] = res
 
    return dp[k][m]
 
# Function to find count of sequences of K
# elements in the range [1, n] where every
# element is a multiple of the previous one
def countSequence(N, K):
 
    # Variable to store
    # the total count
    ans = 0
 
    # Iterate from 1 to N
    for i in range(1, N + 1):
        ans += countSequenceUtil(K, i, N)
 
    # Return ans
    return ans
 
# Driver Code
N = 9
K = 5
 
print(countSequence(N, K))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG {
 
    // Initialize the dp matrix
    static int[, ] dp = new int[1001, 1001];
 
    // Function to find the count of sequences of K
    // elements with first element as m where every
    // element is a multiple of the previous one
    static int countSequenceUtil(int k, int m, int n)
    {
       
        // Base case
        if (k == 1) {
            return 1;
        }
 
        // If the value already exists
        // in the DP then return it
        if (dp[k, m] != -1) {
            return dp[k, m];
        }
 
        // Variable to store the count
        int res = 0;
 
        for (int i = 1; i <= (n / m); i++) {
 
            // Recursive Call
            res += countSequenceUtil(k - 1, m * i, n);
        }
 
        // Store the calculated
        // answer and return it
        return dp[k, m] = res;
    }
 
    // Function to find count of sequences of K
    // elements in the range [1, n] where every
    // element is a multiple of the previous one
    static int countSequence(int N, int K)
    {
 
        // Initializing all values
        // of dp with -1
        for (int i = 0; i < dp.GetLength(0); i++) {
            for (int j = 0; j < dp.GetLength(1); j++) {
                dp[i, j] = -1;
            }
        }
 
        // Variable to store
        // the total count
        int ans = 0;
 
        // Iterate from 1 to N
        for (int i = 1; i <= N; i++) {
            ans += countSequenceUtil(K, i, N);
        }
 
        // Return ans
        return ans;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int N = 9;
        int K = 5;
 
        Console.WriteLine(countSequence(N, K));
    }
}
 
// This code is contributed by ukasp.

Javascript

    // JavaScript implementation for the above approach
 
    // Initialize the dp matrix
    let dp = new Array(1001).fill(-1).map(() => new Array(1001).fill(-1));
 
    // Function to find the count of sequences of K
    // elements with first element as m where every
    // element is a multiple of the previous one
    const countSequenceUtil = (k, m, n) => {
     
        // Base case
        if (k == 1) {
            return 1;
        }
 
        // If the value already exists
        // in the DP then return it
        if (dp[k][m] != -1) {
            return dp[k][m];
        }
 
        // Variable to store the count
        let res = 0;
 
        for (let i = 1; i <= parseInt(n / m); i++) {
 
            // Recursive Call
            res += countSequenceUtil(k - 1,
                m * i, n);
        }
 
        // Store the calculated
        // answer and return it
        return dp[k][m] = res;
    }
 
    // Function to find count of sequences of K
    // elements in the range [1, n] where every
    // element is a multiple of the previous one
    const countSequence = (N, K) => {
 
        // Variable to store
        // the total count
        let ans = 0;
 
        // Iterate from 1 to N
        for (let i = 1; i <= N; i++) {
            ans += countSequenceUtil(K, i, N);
        }
 
        // Return ans
        return ans;
    }
 
    // Driver Code
    let N = 9;
    let K = 5;
 
    document.write(countSequence(N, K));
 
    // This code is contributed by rakeshsahni
 

输出

111

时间复杂度： O(N*K*log N)
辅助空间： O(N*K)