大小为 K 的所有子阵列的 GCD
给定一个大小为 N 的数组arr[]，任务是打印所有大小为 K 的子数组的 GCD 。
示例:
输入: arr[] = {2，4，3，9，14，20，25，17}，K = 2 输出: 2 1 3 1 2 5 1 解释:T8】gcd(2，4}) = 2 gcd(4，3) = 1 gcd(3，9) = 3 gcd(9，14) = 1 gcd(14
输入: arr[] = {2，4，8，24，14，20，25，35，7，49，7}，K = 3 输出:【4 2 2 1 5 1 7 7
方式:思路是生成大小为 K 的所有子阵，打印每个子阵的 GCD。为了有效地计算每个子阵列的 GCD，想法是使用 GCD 的以下属性。
GCD(A 1 ，A 2 ，A 3 ，…，A K ) = GCD(A 1 ，GCD(A 2 ，A 3 ，A 4 ，…。，A K )
按照以下步骤解决问题:
- 初始化一个变量，比如 gcd ，来存储当前子阵列的 GCD 。
- 从给定阵列生成 K 长度的子阵列。
- 应用 GCD 的上述性质，计算每个子阵列的 GCD，并打印得到的结果。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the gcd
// of each subarray of length K
void printSub(int arr[], int N,
int K)
{
for (int i = 0; i <= N - K; i++) {
// Store GCD of subarray
int gcd = arr[i];
for (int j = i + 1; j < i + K;
j++) {
// Update GCD of subarray
gcd = __gcd(gcd, arr[j]);
}
// Print GCD of subarray
cout << gcd << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 2, 4, 3, 9, 14,
20, 25, 17 };
int K = 2;
int N = sizeof(arr)
/ sizeof(arr[0]);
printSub(arr, N, K);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
class GFG{
static int __gcd(int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to print the gcd
// of each subarray of length K
static void printSub(int arr[],
int N, int K)
{
for (int i = 0; i <= N - K; i++)
{
// Store GCD of subarray
int gcd = arr[i];
for (int j = i + 1; j < i + K; j++)
{
// Update GCD of subarray
gcd = __gcd(gcd, arr[j]);
}
// Print GCD of subarray
System.out.print(gcd + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {2, 4, 3, 9,
14, 20, 25, 17};
int K = 2;
int N = arr.length;
printSub(arr, N, K);
}
}
// This code is contributed by Chitranayal
Python 3
# Python3 program to implement
# the above approach
from math import gcd
# Function to print the gcd
# of each subarray of length K
def printSub(arr, N, K):
for i in range(N - K + 1):
# Store GCD of subarray
g = arr[i]
for j in range(i + 1, i + K):
# Update GCD of subarray
g = gcd(g, arr[j])
# Print GCD of subarray
print(g, end = " ")
# Driver Code
if __name__ == '__main__':
arr = [ 2, 4, 3, 9, 14,
20, 25, 17 ]
K = 2
N = len(arr)
printSub(arr, N, K)
# This code is contributed by mohit kumar 29
C
// C# program to implement
// the above approach
using System;
class GFG{
static int __gcd(int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to print the gcd
// of each subarray of length K
static void printSub(int []arr,
int N, int K)
{
for (int i = 0; i <= N - K; i++)
{
// Store GCD of subarray
int gcd = arr[i];
for (int j = i + 1; j < i + K; j++)
{
// Update GCD of subarray
gcd = __gcd(gcd, arr[j]);
}
// Print GCD of subarray
Console.Write(gcd + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {2, 4, 3, 9,
14, 20, 25, 17};
int K = 2;
int N = arr.Length;
printSub(arr, N, K);
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to implement
// the above approach
function __gcd(a, b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to print the gcd
// of each subarray of length K
function prletSub(arr, N, K)
{
for (let i = 0; i <= N - K; i++)
{
// Store GCD of subarray
let gcd = arr[i];
for (let j = i + 1; j < i + K; j++)
{
// Update GCD of subarray
gcd = __gcd(gcd, arr[j]);
}
// Print GCD of subarray
document.write(gcd + " ");
}
}
// Driver Code
let arr = [2, 4, 3, 9,
14, 20, 25, 17];
let K = 2;
let N = arr.length;
prletSub(arr, N, K);
// This code is contributed by avijitmondal1998.
</script>
Output:
2 1 3 1 2 5 1
时间复杂度:O((N–K+1) K)* 辅助空间: O(1)
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