GCD等于1的子阵列数

给定数组arr [] ，任务是查找GCD值等于1的子数组的数量。

例子：

Input: arr[] = {1, 1, 1}
Output: 6
All the subarrays of the given array
will have GCD equal to 1.

Input: arr[] = {2, 2, 2}
Output: 0

方法：主要观察结果是，如果知道子数组arr [l…r]的所有元素的GCD，那么可以获得子数组arr [l…r + 1]的所有元素的GCD通过简单地使用arr [r + 1]来获取先前子数组的GCD即可。
因此，对于每个索引i ，保持向前迭代并计算从索引i到j的GCD并检查其是否等于1 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the required count
int cntSubArr(int* arr, int n)
{
    // To store the final answer
    int ans = 0;
  
    for (int i = 0; i < n; i++) {
  
        // To store the GCD starting from
        // index 'i'
        int curr_gcd = 0;
  
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (int j = i; j < n; j++) {
            curr_gcd = __gcd(curr_gcd, arr[j]);
  
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1);
        }
    }
  
    // Return the final answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 1 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << cntSubArr(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function to return the required count
static int cntSubArr(int []arr, int n)
{
    // To store the final answer
    int ans = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // To store the GCD starting from
        // index 'i'
        int curr_gcd = 0;
  
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (int j = i; j < n; j++) 
        {
            curr_gcd = __gcd(curr_gcd, arr[j]);
  
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1) ? 1 : 0;
        }
    }
  
    // Return the final answer
    return ans;
}
  
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b);     
}
  
// Driver code
public static void main(String []args) 
{
    int arr[] = { 1, 1, 1 };
    int n = arr.length;
  
    System.out.println(cntSubArr(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
from math import gcd
  
# Function to return the required count 
def cntSubArr(arr, n) :
  
    # To store the final answer 
    ans = 0; 
  
    for i in range(n) :
  
        # To store the GCD starting from 
        # index 'i' 
        curr_gcd = 0; 
  
        # Loop to find the gcd of each subarray 
        # from arr[i] to arr[i...n-1] 
        for j in range(i, n) :
            curr_gcd = gcd(curr_gcd, arr[j]); 
  
            # Increment the count if curr_gcd = 1 
            ans += (curr_gcd == 1);
  
    # Return the final answer 
    return ans; 
  
# Driver code 
if __name__ == "__main__" :
  
    arr = [ 1, 1, 1 ]; 
    n = len(arr); 
  
    print(cntSubArr(arr, n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
                      
class GFG 
{
  
// Function to return the required count
static int cntSubArr(int []arr, int n)
{
    // To store the final answer
    int ans = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // To store the GCD starting from
        // index 'i'
        int curr_gcd = 0;
  
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (int j = i; j < n; j++) 
        {
            curr_gcd = __gcd(curr_gcd, arr[j]);
  
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1) ? 1 : 0;
        }
    }
  
    // Return the final answer
    return ans;
}
  
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b);     
}
  
// Driver code
public static void Main(String []args) 
{
    int []arr = { 1, 1, 1 };
    int n = arr.Length;
  
    Console.WriteLine(cntSubArr(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

输出：

6

时间复杂度： O(N ² )