求除 K 的倍数之外的所有 Array 元素的 GCD

给定一个包含N个整数的数组arr[]和一个整数K ，任务是找到除可被K整除的元素之外的所有数组元素的 GCD。

例子：

Input: arr[] = {2, 4, 6, 8, 10}, N = 5, K = 10
Output: 2
Explanation: The multiple of K is 10.
Remaining elements are 2, 4, 6 and 8 the gcd of which are 2.

Input: arr[] ={45, 15, 90, 20, 10}, N = 5, K = 15
Output: 10

方法：这个问题的基本方法是先找到K的倍数，然后找到剩余元素的GCD。

请按照以下步骤解决此问题：

• 遍历数组的所有元素
• 存储一个数组中所有不能被 K 整除的元素。
• 然后对于剩余的元素，找到所有数字的 GCD。
• 返回 GCD 值。

下面是上述方法的实现：

C++

// C++ code to implement the approach
 
#include 
using namespace std;
 
// Function to print GCD
int findPrime(int arr[], int n, int k)
{
    int i, hcf;
    int c = 0;
 
    // Vector used to store all the elements
    // which are not divisible by k
    vector v;
 
    for (i = 0; i < n; i++) {
        if (arr[i] % k == 0)
            continue;
        else
            v.push_back(arr[i]);
    }
    if (v.size() == 0) {
        cout << "NO";
    }
    else if (v.size() == 1) {
        hcf = v[0];
    }
    else {
         
        // Finding the gcd using built in
        // funtion __gcd(value1,value2);
        hcf = __gcd(v[0], v[1]);
        for (i = 2; i < v.size(); i++) {
            hcf = __gcd(arr[i], hcf);
        }
    }
    return hcf;
}
 
// Driver Code
int main()
{
    int N = 5;
    int K = 10;
    int arr[] = { 2, 4, 6, 8, 10 };
     
    // Function call
    cout << findPrime(arr, N, K);
    return 0;
}

Python3

# Python code to implement the approach
 
# Function for GCD
def __gcd(a, b):
    if (b == 0):
        return a
    return __gcd(b, a % b)
 
# Function to print GCD
def findPrime(arr, n, k):
    c = 0
 
    # Vector used to store all the elements
    # which are not divisible by k
    v = []
 
    for i in range(n):
        if (arr[i] % k == 0):
            continue
        else:
            v.append(arr[i])
 
    if (len(v) == 0):
        print("NO")
    elif (len(v) == 1):
        hcf = v[0]
    else:
 
        # Finding the gcd using built in
        # funtion __gcd(value1,value2);
        hcf = __gcd(v[0], v[1])
        for i in range(2, len(v)):
            hcf = __gcd(arr[i], hcf)
 
    return hcf
 
# Driver Code
N = 5
K = 10
arr = [2, 4, 6, 8, 10]
 
# Function call
print(findPrime(arr, N, K))
 
# This code is contributed by shinjanpatra

Javascript

    // JavaScript code to implement the approach
 
    // Functor for GCD
    const __gcd = (a, b) => {
        if (b == 0) return a;
        return __gcd(b, a % b);
    }
    // Function to print GCD
    const findPrime = (arr, n, k) => {
        let i, hcf;
        let c = 0;
 
        // Vector used to store all the elements
        // which are not divisible by k
        let v = [];
 
        for (i = 0; i < n; i++) {
            if (arr[i] % k == 0)
                continue;
            else
                v.push(arr[i]);
        }
        if (v.length == 0) {
            document.write("NO");
        }
        else if (v.length == 1) {
            hcf = v[0];
        }
        else {
 
            // Finding the gcd using built in
            // funtion __gcd(value1,value2);
            hcf = __gcd(v[0], v[1]);
            for (i = 2; i < v.length; i++) {
                hcf = __gcd(arr[i], hcf);
            }
        }
        return hcf;
    }
 
    // Driver Code
 
    let N = 5;
    let K = 10;
    let arr = [2, 4, 6, 8, 10];
 
    // Function call
    document.write(findPrime(arr, N, K));
 
// This code is contributed by rakeshsahni
 

输出

2

时间复杂度：O(N * log M)，其中 M 是数组的最大元素
辅助空间：O(N)