原文:
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
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
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
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
复杂度分析:
- 时间复杂度: o(n)
- 空间复杂度: o(n)
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