给定一个n个整数的数组arr [] 。任务是从数组K中找到一个元素，使得
- K是素数。
- 并且，在所有可能的K值中，所有有效i的arr [i]％K均为最大值
如果数组中没有素数，则输出-1 。
例子：
Input: arr[] = {2, 10, 15, 7, 6, 8, 13}
Output: 13
2, 7 and 13 are the only prime numbers in the array.
The maximum possible value of arr[i] % 2 is 1 i.e. 15 % 2 = 1.
For 7, it is 6 % 7 = 6
For 13, 10 % 13 = 10 (Maximum possible)
Input: arr[] = {23, 13, 6, 2, 15, 18, 8}
Output: 23
方法：为了最大化arr [i]％K的值， K必须是数组中的最大素数，而arr [i]必须是数组中小于K的最大元素。因此，该问题现在简化为从数组中找到最大素数。为了做到这一点，
- 首先使用Sieve从数组中找到所有小于或等于最大元素的质数。
- 然后，从数组中找到最大素数并打印。如果数组中没有素数，则打印-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
// prime number from the array
int getPrime(int arr[], int n)
{
// Find maximum value in the array
int max_val = *max_element(arr, arr + n);
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
vector prime(max_val + 1, true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// To store the maximum prime number
int maximum = -1;
for (int i = 0; i < n; i++) {
// If current element is prime
// then update the maximum prime
if (prime[arr[i]])
maximum = max(maximum, arr[i]);
}
// Return the maximum prime
// number from the array
return maximum;
}
// Driver code
int main()
{
int arr[] = { 2, 10, 15, 7, 6, 8, 13 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << getPrime(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the required
// prime number from the array
static int getPrime(int arr[], int n)
{
// Find maximum value in the array
int max_val = Arrays.stream(arr).max().getAsInt();
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
Vector prime = new Vector<>(max_val + 1);
for(int i = 0; i < max_val + 1; i++)
prime.add(i,Boolean.TRUE);
// Remaining part of SIEVE
prime.add(1,Boolean.FALSE);
prime.add(2,Boolean.FALSE);
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed, then
// it is a prime
if (prime.get(p) == true)
{
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime.add(i,Boolean.FALSE);
}
}
// To store the maximum prime number
int maximum = -1;
for (int i = 0; i < n; i++)
{
// If current element is prime
// then update the maximum prime
if (prime.get(arr[i]))
{
maximum = Math.max(maximum, arr[i]);
}
}
// Return the maximum prime
// number from the array
return maximum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 10, 15, 7, 6, 8, 13 };
int n = arr.length;
System.out.println(getPrime(arr, n));
}
}
// This code has been contributed by 29AjayKumar Python3
# Python 3 implementation of the approach
from math import sqrt
# Function to return the required
# prime number from the array
def getPrime(arr, n):
# Find maximum value in the array
max_val = arr[0]
for i in range(len(arr)):
# USE SIEVE TO FIND ALL PRIME NUMBERS LESS
# THAN OR EQUAL TO max_val
# Create a boolean array "prime[0..n]". A
# value in prime[i] will finally be false
# if i is Not a prime, else true.
if(arr[i] > max_val):
max_val = arr[i]
prime = [True for i in range(max_val + 1)]
# Remaining part of SIEVE
prime[0] = False
prime[1] = False
for p in range(2, int(sqrt(max_val)) + 1, 1):
# If prime[p] is not changed, then
# it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * 2, max_val + 1, p):
prime[i] = False
# To store the maximum prime number
maximum = -1
for i in range(n):
# If current element is prime
# then update the maximum prime
if (prime[arr[i]]):
maximum = max(maximum, arr[i])
# Return the maximum prime
# number from the array
return maximum
# Driver code
if __name__ == '__main__':
arr = [2, 10, 15, 7, 6, 8, 13]
n = len(arr)
print(getPrime(arr, n))
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
// Function to return the required
// prime number from the array
static int getPrime(int []arr, int n)
{
// Find maximum value in the array
int max_val = arr.Max();
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
List prime = new List(max_val + 1);
for(int i = 0; i < max_val + 1; i++)
prime.Insert(i, true);
// Remaining part of SIEVE
prime.Insert(1, false);
prime.Insert(2, false);
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * 2;
i <= max_val; i += p)
prime.Insert(i, false);
}
}
// To store the maximum prime number
int maximum = -1;
for (int i = 0; i < n; i++)
{
// If current element is prime
// then update the maximum prime
if (prime[arr[i]])
{
maximum = Math.Max(maximum, arr[i]);
}
}
// Return the maximum prime
// number from the array
return maximum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 10, 15, 7, 6, 8, 13 };
int n = arr.Length;
Console.WriteLine(getPrime(arr, n));
}
}
// This code contributed by Rajput-Ji PHP
输出：
13