除1和数字本身以外的数组中每个元素的最大除数

给定N个整数的数组arr [] ，任务是为数组中除1和数字本身以外的每个元素找到最大的除数。如果没有这样的除数，则打印-1。

例子：

Input: arr[] = {5, 6, 7, 8, 9, 10}
Output: -1 3 -1 4 3 5
Divisors(5) = {1, 5}
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1
Divisors(6) = [1, 2, 3, 6]
-> largest divisor other than 1 and the number itself = 3
Divisors(7) = [1, 7]
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1
Divisors(8) = [1, 2, 4, 8]
-> largest divisor other than 1 and the number itself = 4
Divisors(9) = [1, 3, 9]
-> largest divisor other than 1 and the number itself = 3
Divisors(10) = [1, 2, 5, 10]
-> largest divisor other than 1 and the number itself = 5

Input: arr[] = {15, 16, 17, 18, 19, 20, 21}
Output: 5 8 -1 9 -1 10 7

天真的方法：想法是遍历所有数组元素，并使用本文讨论的方法为每个元素找到最大的除数。

时间复杂度：O(N *＆Sqrt; N)

高效的方法：更好的解决方案是预先计算2到10 ^5之间的数字的最大除数，然后对数组运行循环并打印预先计算的答案。

• 使用Eratosthenes筛子标记质数，并存储每个数的最小质数。
• 现在，任何数字的最大除数将是number / minimum_prime_divisor 。
• 使用预先计算的答案为每个数字找到最大的除数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
  
#include 
using namespace std;
  
#define int long long
const int maxin = 100001;
  
// Divisors array to keep track
// of the maximum divisor
int divisors[maxin];
  
// Function to pre-compute the prime
// numbers and largest divisors
void Calc_Max_Div(int arr[], int n)
{
  
    // Visited array to keep
    // track of prime numbers
    bool vis[maxin];
    memset(vis, 1, maxin);
  
    // 0 and 1 are not prime numbers
    vis[0] = vis[1] = 0;
  
    // Initialising divisors[i] = i
    for (int i = 1; i <= maxin; i++)
        divisors[i] = i;
  
    // For all the numbers divisible by 2
    // the maximum divisor will be number / 2
    for (int i = 4; i <= maxin; i += 2) {
        vis[i] = 0;
        divisors[i] = i / 2;
    }
    for (int i = 3; i <= maxin; i += 2) {
  
        // If divisors[i] is not equal to i then
        // this means that divisors[i] contains
        // minimum prime divisor for the number
        if (divisors[i] != i) {
  
            // Update the answer to
            // i / smallest_prime_divisor[i]
            divisors[i] = i / divisors[i];
        }
  
        // Condition if i is a prime number
        if (vis[i] == 1) {
            for (int j = i * i; j < maxin; j += i) {
                vis[j] = 0;
  
                // If divisors[j] is equal to j then
                // this means that i is the first prime
                // divisor for j so we update divi[j] = i
                if (divisors[j] == j)
                    divisors[j] = i;
            }
        }
    }
  
    for (int i = 0; i < n; i++) {
  
        // If the current element is prime
        // then it has no divisors
        // other than 1 and itself
        if (divisors[arr[i]] == arr[i])
            cout << "-1 ";
        else
            cout << divisors[arr[i]] << " ";
    }
}
  
// Driver code
int32_t main()
{
    int arr[] = { 5, 6, 7, 8, 9, 10 };
    int n = sizeof(arr) / sizeof(int);
  
    Calc_Max_Div(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG 
{
      
    final static int maxin = 10001; 
      
    // Divisors array to keep track 
    // of the maximum divisor 
    static int divisors[] = new int[maxin + 1]; 
      
    // Function to pre-compute the prime 
    // numbers and largest divisors 
    static void Calc_Max_Div(int arr[], int n) 
    { 
      
        // Visited array to keep 
        // track of prime numbers 
        int vis[] = new int[maxin + 1];
          
        for(int i = 0;i 

---

Python3

# Python3 implementation of the approach 
maxin = 100001; 
  
# Divisors array to keep track 
# of the maximum divisor 
divisors = [0] * (maxin + 1); 
  
# Function to pre-compute the prime 
# numbers and largest divisors 
def Calc_Max_Div(arr, n) :
  
    # Visited array to keep 
    # track of prime numbers 
    vis = [1] * (maxin + 1); 
  
    # 0 and 1 are not prime numbers 
    vis[0] = vis[1] = 0; 
  
    # Initialising divisors[i] = i 
    for i in range(1, maxin + 1) :
        divisors[i] = i; 
  
    # For all the numbers divisible by 2 
    # the maximum divisor will be number / 2 
    for i in range(4 , maxin + 1, 2) :
        vis[i] = 0; 
        divisors[i] = i // 2; 
      
    for i in range(3, maxin + 1, 2) :
  
        # If divisors[i] is not equal to i then 
        # this means that divisors[i] contains 
        # minimum prime divisor for the number 
        if (divisors[i] != i) :
  
            # Update the answer to 
            # i / smallest_prime_divisor[i] 
            divisors[i] = i // divisors[i]; 
      
        # Condition if i is a prime number 
        if (vis[i] == 1) :
            for j in range( i * i, maxin, i) : 
                vis[j] = 0; 
  
                # If divisors[j] is equal to j then 
                # this means that i is the first prime 
                # divisor for j so we update divi[j] = i 
                if (divisors[j] == j) :
                    divisors[j] = i; 
          
    for i in range(n) :
  
        # If the current element is prime 
        # then it has no divisors 
        # other than 1 and itself 
        if (divisors[arr[i]] == arr[i]) :
            print("-1 ", end = ""); 
        else :
            print(divisors[arr[i]], end = " "); 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 5, 6, 7, 8, 9, 10 ]; 
    n = len(arr); 
  
    Calc_Max_Div(arr, n); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach 
using System;
  
class GFG 
{
      
    static int maxin = 10001; 
      
    // Divisors array to keep track 
    // of the maximum divisor 
    static int []divisors = new int[maxin + 1]; 
      
    // Function to pre-compute the prime 
    // numbers and largest divisors 
    static void Calc_Max_Div(int []arr, int n) 
    { 
      
        // Visited array to keep 
        // track of prime numbers 
        int []vis = new int[maxin + 1];
          
        for(int i = 0; i < maxin + 1 ; i++)
            vis[i] = 1;
  
        // 0 and 1 are not prime numbers 
        vis[0] = vis[1] = 0; 
      
        // Initialising divisors[i] = i 
        for (int i = 1; i <= maxin; i++) 
            divisors[i] = i; 
      
        // For all the numbers divisible by 2 
        // the maximum divisor will be number / 2 
        for (int i = 4; i <= maxin; i += 2) 
        { 
            vis[i] = 0; 
            divisors[i] = i / 2; 
        } 
        for (int i = 3; i <= maxin; i += 2) 
        { 
      
            // If divisors[i] is not equal to i then 
            // this means that divisors[i] contains 
            // minimum prime divisor for the number 
            if (divisors[i] != i)
            { 
      
                // Update the answer to 
                // i / smallest_prime_divisor[i] 
                divisors[i] = i / divisors[i]; 
            } 
      
            // Condition if i is a prime number 
            if (vis[i] == 1) 
            { 
                for (int j = i * i; j < maxin; j += i)
                { 
                    vis[j] = 0; 
      
                    // If divisors[j] is equal to j then 
                    // this means that i is the first prime 
                    // divisor for j so we update divi[j] = i 
                    if (divisors[j] == j) 
                        divisors[j] = i; 
                } 
            } 
        } 
      
        for (int i = 0; i < n; i++) 
        { 
      
            // If the current element is prime 
            // then it has no divisors 
            // other than 1 and itself 
            if (divisors[arr[i]] == arr[i]) 
                    Console.Write("-1 "); 
            else
                    Console.Write(divisors[arr[i]] + " "); 
        } 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int []arr = { 5, 6, 7, 8, 9, 10 }; 
        int n = arr.Length; 
      
        Calc_Max_Div(arr, n); 
    } 
}
  
// This code is contributed by AnkitRai01

输出：

-1 3 -1 4 3 5

时间复杂度：O(N)