数组中除 1 和数字本身之外的每个元素的最大除数

给定一个由N 个整数组成的数组arr[] ，任务是找到数组中每个元素的最大除数，而不是 1 和数字本身。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 
 

朴素的方法：这个想法是迭代所有数组元素，并使用本文中讨论的方法为每个元素找到最大的除数。
时间复杂度：O(N * √N)

有效的方法：更好的解决方案是预先计算从 2 到 10 ⁵的数字的最大除数，然后为数组运行一个循环并打印预先计算的答案。

• 使用 Sieve of Eratosthenes 标记素数并存储每个数的最小素数除数。
• 现在任何数字的最大除数将是number /smallest_prime_divisor 。
• 使用预先计算的答案找到每个数字的最大除数。

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

Javascript

 
// Javascript implementation of the approach
var maxin = 100001;
 
// Divisors array to keep track
// of the maximum divisor
var divisors = Array(maxin).fill(0);
 
// Function to pre-compute the prime
// numbers and largest divisors
function Calc_Max_Div(arr, n)
{
 
    // Visited array to keep
    // track of prime numbers
    var vis = Array(maxin).fill(1);
 
    // 0 and 1 are not prime numbers
    vis[0] = vis[1] = 0;
      
    var i,j;
     
    // Initialising divisors[i] = i
    for(i = 1; i <= maxin; i++)
        divisors[i] = i;
 
    // For all the numbers divisible by 2
    // the maximum divisor will be number / 2
    for(i = 4; i <= maxin; i += 2)
    {
        vis[i] = 0;
        divisors[i] = i / 2;
    }
    for(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(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(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])
            document.write("-1 ");
        else
            document.write(divisors[arr[i]] + " ");
    }
}
 
// Driver code
var arr = [ 5, 6, 7, 8, 9, 10 ];
var n = arr.length;
 
Calc_Max_Div(arr, n);
 
// This code is contributed by bgangwar59
 

输出：

-1 3 -1 4 3 5

时间复杂度：O(N)

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