给定长度为N的数组arr [] ，任务是找到具有最大可能GCD值的最长子数组的长度。
例子：
Input: arr[] = {1, 2, 2}
Output: 2
Here all possible sub-arrays and there GCD’s are:
1) {1} -> 1
2) {2} -> 2
3) {2} -> 2
4) {1, 2} -> 1
5) {2, 2} -> 2
6) {1, 2, 3} -> 1
Here, the maximum GCD value is 2 and longest sub-array having GCD = 2 is {2, 2}.
Thus, the answer is {2, 2}.
Input: arr[] = {3, 3, 3, 3}
Output: 4
天真的方法：生成所有可能的子阵列，并分别找到每个子阵列的GCD，以找到最长的子阵列。该方法将花费O(N 3 )时间来解决该问题。
更好的方法：最大GCD值将始终等于阵列中存在的最大数字。假设数组中存在的最大数字是X。现在，任务是找到具有所有X的最大子数组。使用两指针方法可以完成相同的操作。下面是算法：
- 在数组中找到最大的数字。让我们将此数字称为X。
- 从i = 0开始运行循环
- 如果arr [i]！= X，则增加i并继续。
- 否则初始化j = i 。
- 当j
- 将答案更新为ans = max(ans，j – i) 。
- 更新我为我= j的。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the length of
// the largest subarray with
// maximum possible GCD
int findLength(int* arr, int n)
{
// To store the maximum number
// present in the array
int x = 0;
// Finding the maximum element
for (int i = 0; i < n; i++)
x = max(x, arr[i]);
// To store the final answer
int ans = 0;
// Two pointer
for (int i = 0; i < n; i++) {
if (arr[i] != x)
continue;
// Running a loop from j = i
int j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
j++;
// Updating the answer
ans = max(ans, j - i);
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 2 };
int n = sizeof(arr) / sizeof(int);
cout << findLength(arr, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return the length of
// the largest subarray with
// maximum possible GCD
static int findLength(int []arr, int n)
{
// To store the maximum number
// present in the array
int x = 0;
// Finding the maximum element
for (int i = 0; i < n; i++)
x = Math.max(x, arr[i]);
// To store the final answer
int ans = 0;
// Two pointer
for (int i = 0; i < n; i++)
{
if (arr[i] != x)
continue;
// Running a loop from j = i
int j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
{
j++;
if (j >= n )
break;
}
// Updating the answer
ans = Math.max(ans, j - i);
}
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 2 };
int n = arr.length;
System.out.println(findLength(arr, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
# Function to return the length of
# the largest subarray with
# maximum possible GCD
def findLength(arr, n) :
# To store the maximum number
# present in the array
x = 0;
# Finding the maximum element
for i in range(n) :
x = max(x, arr[i]);
# To store the final answer
ans = 0;
# Two pointer
for i in range(n) :
if (arr[i] != x) :
continue;
# Running a loop from j = i
j = i;
# Condition for incrementing 'j'
while (arr[j] == x) :
j += 1;
if j >= n :
break
# Updating the answer
ans = max(ans, j - i);
return ans;
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 2 ];
n = len(arr);
print(findLength(arr, n));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the length of
// the largest subarray with
// maximum possible GCD
static int findLength(int []arr, int n)
{
// To store the maximum number
// present in the array
int x = 0;
// Finding the maximum element
for (int i = 0; i < n; i++)
x = Math.Max(x, arr[i]);
// To store the final answer
int ans = 0;
// Two pointer
for (int i = 0; i < n; i++)
{
if (arr[i] != x)
continue;
// Running a loop from j = i
int j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
{
j++;
if (j >= n )
break;
}
// Updating the answer
ans = Math.Max(ans, j - i);
}
return ans;
}
// Driver code
public static void Main ()
{
int []arr = { 1, 2, 2 };
int n = arr.Length;
Console.WriteLine(findLength(arr, n));
}
}
// This code is contributed by AnkitRai01输出：
2