给定大小为N的数组arr [] ,任务是打印大小为K的所有子数组的GCD。
例子:
Input: arr[] = {2, 4, 3, 9, 14, 20, 25, 17}, K = 2
Output: 2 1 3 1 2 5 1
Explanation:
gcd(2, 4}) = 2
gcd(4, 3) = 1
gcd(3, 9) = 3
gcd(9, 14) = 1
gcd(14, 20) = 2
gcd(20, 25) = 5
gcd(25, 17) = 1
Therefore, the required output is {2, 1, 3, 1, 2, 5, 1}
Input: arr[] = {2, 4, 8, 24, 14, 20, 25, 35, 7, 49, 7}, K = 3
Output: 2 4 2 2 1 5 1 7 7
方法:想法是生成所有大小为K的子数组,并打印每个子数组的GCD。为了有效地计算每个子阵列的GCD,其思想是使用GCD的以下属性。
GCD(A1, A2, A3, …, AK) = GCD(A1, GCD(A2, A3, A4, …., AK))
请按照以下步骤解决问题:
- 初始化一个变量,例如gcd ,以存储当前子数组的GCD 。
- 从给定数组生成K个长度的子数组。
- 应用GCD的上述属性,计算每个子数组的GCD,然后打印获得的结果。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
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
// 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
Python3
# 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
输出:
2 1 3 1 2 5 1
时间复杂度: O((N – K + 1)* K)
辅助空间: O(1)