打印大小为N的序列，其中每个项都是前K个项的和

给定两个整数N和K ，任务是生成一系列N个项，其中每个项都是前K个项的和。
注意：系列的第一项为1 ，如果先前的项不足，则其他项应为0 。
例子：

Input: N = 8, K = 3 
Output: 1 1 2 4 7 13 24 44 
Explanation: 
Series is generated as follows: 
a[0] = 1 
a[1] = 1 + 0 + 0 = 1 
a[2] = 1 + 1 + 0 = 2 
a[3] = 2 + 1 + 1 = 4 
a[4] = 4 + 2 + 1 = 7 
a[5] = 7 + 4 + 2 = 13 
a[6] = 13 + 7 + 4 = 24 
a[7] = 24 + 13 + 7 = 44
Input: N = 10, K = 4 
Output: 1 1 2 4 8 15 29 56 108 208 

天真的方法：这个想法是运行两个循环以生成N个系列项。下面是步骤说明：

• 从0到N – 1遍历第一个循环，以生成序列的每个项。
• 运行从max(0，i – K)到i的循环，以计算前K个项的总和。
• 将总和更新为当前系列的当前索引。

下面是上述方法的实现：

C++

// C++ implementation to find the
// series in which every term is
// sum of previous K terms
 
#include 
 
using namespace std;
 
// Function to generate the
// series in the form of array
void sumOfPrevK(int N, int K)
{
    int arr[N];
    arr[0] = 1;
 
    // Pick a starting point
    for (int i = 1; i < N; i++) {
        int j = i - 1, count = 0,
            sum = 0;
        // Find the sum of all
        // elements till count < K
        while (j >= 0 && count < K) {
            sum += arr[j];
            j--;
            count++;
        }
        // Find the value of
        // sum at i position
        arr[i] = sum;
    }
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 10, K = 4;
    sumOfPrevK(N, K);
    return 0;
}

Java

// Java implementation to find the
// series in which every term is
// sum of previous K terms
 
class Sum {
    // Function to generate the
    // series in the form of array
    void sumOfPrevK(int N, int K)
    {
        int arr[] = new int[N];
        arr[0] = 1;
 
        // Pick a starting point
        for (int i = 1; i < N; i++) {
            int j = i - 1, count = 0,
                sum = 0;
            // Find the sum of all
            // elements till count < K
            while (j >= 0 && count < K) {
                sum += arr[j];
                j--;
                count++;
            }
            // Find the value of
            // sum at i position
            arr[i] = sum;
        }
        for (int i = 0; i < N; i++) {
            System.out.print(arr[i] + " ");
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        Sum s = new Sum();
        int N = 10, K = 4;
        s.sumOfPrevK(N, K);
    }
}

Python3

# Python3 implementation to find the
# series in which every term is
# sum of previous K terms
 
# Function to generate the
# series in the form of array
def sumOfPrevK(N, K):
    arr = [0 for i in range(N)]
    arr[0] = 1
 
    # Pick a starting point
    for i in range(1,N):
        j = i - 1
        count = 0
        sum = 0
         
        # Find the sum of all
        # elements till count < K
        while (j >= 0 and count < K):
            sum = sum + arr[j]
            j = j - 1
            count = count + 1
 
        # Find the value of
        # sum at i position
        arr[i] = sum
 
    for i in range(0, N):
        print(arr[i])
 
# Driver Code
N = 10
K = 4
sumOfPrevK(N, K)
 
# This code is contributed by Sanjit_Prasad

C#

// C# implementation to find the
// series in which every term is
// sum of previous K terms
using System;
 
class Sum {
    // Function to generate the
    // series in the form of array
    void sumOfPrevK(int N, int K)
    {
        int []arr = new int[N];
        arr[0] = 1;
  
        // Pick a starting point
        for (int i = 1; i < N; i++) {
            int j = i - 1, count = 0,
                sum = 0;
 
            // Find the sum of all
            // elements till count < K
            while (j >= 0 && count < K) {
                sum += arr[j];
                j--;
                count++;
            }
 
            // Find the value of
            // sum at i position
            arr[i] = sum;
        }
        for (int i = 0; i < N; i++) {
            Console.Write(arr[i] + " ");
        }
    }
  
    // Driver Code
    public static void Main(String []args)
    {
        Sum s = new Sum();
        int N = 10, K = 4;
        s.sumOfPrevK(N, K);
    }
}
 
// This code is contributed by 29AjayKumar

性能分析：

• 时间复杂度： O(N * K)
• 空间复杂度： O(N)

高效方法：想法是将当前总和存储在变量中，并在每个步骤中减去最后的^第K^个项，并将最后一个项添加到预和中，以计算序列的每个项。
下面是上述方法的实现：

C++

// C++ implementation to find the
// series in which every term is
// sum of previous K terms
 
#include 
 
using namespace std;
 
// Function to generate the
// series in the form of array
void sumOfPrevK(int N, int K)
{
    int arr[N], prevsum = 0;
    arr[0] = 1;
 
    // Pick a starting point
    for (int i = 0; i < N - 1; i++) {
 
        // Computing the previous sum
        if (i < K) {
            arr[i + 1] = arr[i] + prevsum;
            prevsum = arr[i + 1];
        }
        else {
            arr[i + 1] = arr[i] + prevsum
                         - arr[i + 1 - K];
            prevsum = arr[i + 1];
        }
    }
 
    // Loop to print the series
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 8, K = 3;
    sumOfPrevK(N, K);
    return 0;
}

Java

// Java implementation to find the
// series in which every term is
// sum of previous K terms
 
class Sum {
    // Function to generate the
    // series in the form of array
    void sumOfPrevK(int N, int K)
    {
        int arr[] = new int[N];
        int prevsum = 0;
        arr[0] = 1;
 
        // Pick a starting point
        for (int i = 0; i < N - 1; i++) {
 
            // Computing the previous sum
            if (i < K) {
                arr[i + 1] = arr[i] + prevsum;
                prevsum = arr[i + 1];
            }
            else {
                arr[i + 1] = arr[i] + prevsum
                             - arr[i + 1 - K];
                prevsum = arr[i + 1];
            }
        }
 
        // Loop to print the series
        for (int i = 0; i < N; i++) {
            System.out.print(arr[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        Sum s = new Sum();
        int N = 8, K = 3;
        s.sumOfPrevK(N, K);
    }
}

Python3

# Python3 implementation to find the
# series in which every term is
# sum of previous K terms
 
# Function to generate the
# series in the form of array
def sumOfPrevK(N, K):
    arr = [0]*N;
    prevsum = 0;
    arr[0] = 1;
 
    # Pick a starting point
    for i in range(N-1):
 
        # Computing the previous sum
        if (i < K):
            arr[i + 1] = arr[i] + prevsum;
            prevsum = arr[i + 1];
        else:
            arr[i + 1] = arr[i] + prevsum - arr[i + 1 - K];
            prevsum = arr[i + 1];
 
    # Loop to print the series
    for i in range(N):
        print(arr[i], end=" ");
     
# Driver code
if __name__ == '__main__':
    N = 8;
    K = 3;
    sumOfPrevK(N, K);
         
# This code is contributed by 29AjayKumar

C＃

// C# implementation to find the
// series in which every term is
// sum of previous K terms
using System;
public class Sum {
    // Function to generate the
    // series in the form of array
    void sumOfPrevK(int N, int K)
    {
        int []arr = new int[N];
        int prevsum = 0;
        arr[0] = 1;
  
        // Pick a starting point
        for (int i = 0; i < N - 1; i++) {
  
            // Computing the previous sum
            if (i < K) {
                arr[i + 1] = arr[i] + prevsum;
                prevsum = arr[i + 1];
            }
            else {
                arr[i + 1] = arr[i] + prevsum
                             - arr[i + 1 - K];
                prevsum = arr[i + 1];
            }
        }
  
        // Loop to print the series
        for (int i = 0; i < N; i++) {
            Console.Write(arr[i] + " ");
        }
    }
  
    // Driver code
    public static void Main(String []args)
    {
        Sum s = new Sum();
        int N = 8, K = 3;
        s.sumOfPrevK(N, K);
    }
}
 
// This code is contributed by 29AjayKumar

复杂度分析：

• 时间复杂度： O(N)
• 空间复杂度： O(N)