打印大小为 N 的序列，其中每一项都是前 K 项的总和
原文:https://www . geesforgeks . org/print-the-sequence-of-size-n-in-每个术语都是前 k 个术语的总和/
给定两个整数 N 和 K ，任务是生成一系列 N 项，其中每个项都是之前的 K 项的和。 注:该系列的第一个术语是 1 ，如果之前的术语不够，那么其他术语应该是 0 。 示例:
输入: N = 8， K = 3 输出: 1 1 2 4 7 13 24 44 说明: 系列生成如下: 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]= 1 = 13+7+4 = 24 a[7]= 24+13+7 = 44 输入: N = 10，K = 4 输出:1 2 4 8 15 29 56 108 208
天真方法:想法是运行两个循环生成 N 项级数。下面是步骤的图示:
- 遍历从 0 到N–1的第一个循环，生成该系列的每个项。
- 运行从 max(0，I–K)到 i 的循环，计算前 K 项的总和。
- 将当前系列索引的总和更新为当前术语。
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 <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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);
}
}
Python 3
# 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
java 描述语言
<script>
// JavaScript 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
function sumOfPrevK(N, K)
{
let arr = new Array(N);
arr[0] = 1;
// Pick a starting point
for (let i = 1; i < N; i++) {
let 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 (let i = 0; i < N; i++) {
document.write(arr[i] + " ");
}
}
// Driver Code
let N = 10, K = 4;
sumOfPrevK(N, K);
// This code is contributed by _saurabh_jaiswal
</script>
输出:
1 1 2 4 8 15 29 56 108 208
性能分析:
- 时间复杂度: O(N * K)
- 空间复杂度: O(N)
有效方法:想法是将当前的和存储在一个变量中，在每一步中减去最后一个 K 第T5】项，并将最后一项添加到预和中，以计算该系列的每个项。 以下是上述方法的实现:
C++
// C++ implementation to find the
// series in which every term is
// sum of previous K terms
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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);
}
}
Python 3
# 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
java 描述语言
<script>
// Javascript 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
function sumOfPrevK(N, K)
{
let arr = new Array(N), prevsum = 0;
arr[0] = 1;
// Pick a starting point
for (let 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 (let i = 0; i < N; i++) {
document.write(arr[i] + " ");
}
}
// Driver Code
let N = 8, K = 3;
sumOfPrevK(N, K);
// This code is contributed by rishavmahato348.
</script>
复杂度分析:
- 时间复杂度: O(N)
- 空间复杂度: O(N)
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