使用最多 K 的自然数求和到 N 的方法，允许重复

给定两个整数N和K ，任务是找到将N表示为[1, K]范围内正整数之和的方法总数，其中每个整数可以多次选择。

例子：

Input: N = 8, K = 2
Output: 5
Explanation: All possible ways of representing N as sum of positive integers less than or equal to K are:

1. {1, 1, 1, 1, 1, 1, 1, 1}, the sum is 8.
2. {2, 1, 1, 1, 1, 1, 1}, the sum is 8.
3. {2, 2, 1, 1, 1, 1}, the sum is 8.
4. 2, 2, 2, 1, 1}, the sum is 8.
5. {2, 2, 2, 2}}, the sum is 8.

Therefore, the total number of ways is 5.

Input: N = 2, K = 2
Output: 2

朴素方法：解决给定问题的最简单方法是生成在[1, K]范围内选择整数的所有可能组合，并计算总和为N 的那些组合。

时间复杂度： O(K ^N )
辅助空间： O(1)

高效方法：上述方法具有重叠子问题和最优子结构。因此，为了优化，需要基于以下观察进行动态规划：

• 考虑到dp[i]将表示i的方式总数存储为[1, K]范围内整数的总和，那么状态转换可以定义为：
  • 对于[1, K]范围内的i和[1, N]范围内的每个j
  • dp[j] 的值等于(dp[j]+ dp[j – i]) ，对于所有j ≥ i 。

请按照以下步骤解决问题：

• 初始化一个数组，比如dp[] ，所有元素都为0 ，以存储所有递归状态。
• 将dp[0]初始化为1 。
• 现在，使用变量i迭代范围[1, K]并执行以下步骤：
  • 迭代范围[1, N] ，使用变量j ，并将dp[j]的值更新为dp[j]+ dp[j – i] ，如果j ≥ i 。
• 完成以上步骤后，打印dp[N]的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
int NumberOfways(int N, int K)
{
 
    // Initialize a list
    vector dp(N + 1, 0);
   
    // Update dp[0] to 1
    dp[0] = 1;
 
    // Iterate over the range [1, K + 1]
    for (int row = 1; row < K + 1; row++)
    {
 
        // Iterate over the range [1, N + 1]
        for (int col = 1; col < N + 1; col++)
        {
 
            // If col is greater
            // than or equal to row
            if (col >= row)
               
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }
 
    // Return the total number of ways
    return(dp[N]);
}
 
// Driver Code
int main()
{
  int N = 8;
  int K = 2;
 
  cout << (NumberOfways(N, K));
}
 
// This code is contributed by mohit kumar 29.

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
static int NumberOfways(int N, int K)
{
     
    // Initialize a list
    int[] dp = new int[N + 1];
   
    // Update dp[0] to 1
    dp[0] = 1;
 
    // Iterate over the range [1, K + 1]
    for(int row = 1; row < K + 1; row++)
    {
 
        // Iterate over the range [1, N + 1]
        for(int col = 1; col < N + 1; col++)
        {
             
            // If col is greater
            // than or equal to row
            if (col >= row)
               
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }
 
    // Return the total number of ways
    return(dp[N]);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given inputs
    int N = 8;
    int K = 2;
     
    System.out.println(NumberOfways(N, K));
}
}
 
// This code is contributed by offbeat

Python

# Python program for the above approach
 
# Function to find the total number of
# ways to represent N as the sum of
# integers over the range [1, K]
def NumberOfways(N, K):
   
    # Initialize a list
    dp = [0] * (N + 1)
     
    # Update dp[0] to 1
    dp[0] = 1
     
    # Iterate over the range [1, K + 1]
    for row in range(1, K + 1):
       
        # Iterate over the range [1, N + 1]
        for col in range(1, N + 1):
           
            # If col is greater
            # than or equal to row
            if (col >= row):
               
                # Update current
                # dp[col] state
                dp[col] = dp[col] + dp[col - row]
                 
                 
    # Return the total number of ways
    return(dp[N])
 
# Driver Code
 
N = 8
K = 2
 
print(NumberOfways(N, K))

C#

// C# program for the above approach
using System;
class GFG
{
   
    // Function to find the total number of
    // ways to represent N as the sum of
    // integers over the range [1, K]
    static int NumberOfways(int N, int K)
    {
 
        // Initialize a list
        int[] dp = new int[(N + 1)];
 
        // Update dp[0] to 1
        dp[0] = 1;
 
        // Iterate over the range [1, K + 1]
        for (int row = 1; row < K + 1; row++) {
 
            // Iterate over the range [1, N + 1]
            for (int col = 1; col < N + 1; col++) {
 
                // If col is greater
                // than or equal to row
                if (col >= row)
 
                    // Update current
                    // dp[col] state
                    dp[col] = dp[col] + dp[col - row];
            }
        }
 
        // Return the total number of ways
        return (dp[N]);
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 8;
        int K = 2;
 
        Console.WriteLine(NumberOfways(N, K));
    }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript implementation for the above approach
 
// Function to find the total number of
// ways to represent N as the sum of
// letegers over the range [1, K]
function NumberOfways(N, K)
{
     
    // Initialize a list
    let dp = Array.from({length: N +1}, (_, i) => 0);
   
    // Update dp[0] to 1
    dp[0] = 1;
 
    // Iterate over the range [1, K + 1]
    for(let row = 1; row < K + 1; row++)
    {
 
        // Iterate over the range [1, N + 1]
        for(let col = 1; col < N + 1; col++)
        {
             
            // If col is greater
            // than or equal to row
            if (col >= row)
               
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }
 
    // Return the total number of ways
    return(dp[N]);
}
 
    // Driver Code
     
    // Given inputs
    let N = 8;
    let K = 2;
     
    document.write(NumberOfways(N, K));
 

输出：

5

时间复杂度： O(N * K)
辅助空间： O(N)