给定一个数组ARR []，任务是找到最大之和给定的阵列中的每个号码的最小素因子。
例子：
Input: arr[] = {15}
Output: 8
The maximum and the minimum prime factors
of 15 are 5 and 3 respectively.
Input: arr[] = {5, 10, 15, 20, 25, 30}
Output: 10 7 8 7 10 7
方法：想法是使用Eratosthenes筛子来预先计算每个数字的所有最小和最大素数因子，并将其存储在两个数组中。预计算之后，可以在恒定时间内找到最小和最大素因数的总和。
下面是上述方法的实现：
CPP
// C++ implementation of the approach
#include
using namespace std;
const int MAX = 100000;
// max_prime[i] represent maximum prime
// number that divides the number i
int max_prime[MAX];
// min_prime[i] represent minimum prime
// number that divides the number i
int min_prime[MAX];
// Function to store the minimum prime factor
// and the maximum prime factor in two arrays
void sieve(int n)
{
for (int i = 2; i <= n; ++i) {
// Check for prime number
// if min_prime[i]>0,
// then it is not a prime number
if (min_prime[i] > 0) {
continue;
}
// if i is a prime number
// min_prime number that divide prime number
// and max_prime number that divide prime number
// is the number itself.
min_prime[i] = i;
max_prime[i] = i;
int j = i + i;
while (j <= n) {
if (min_prime[j] == 0) {
// If this number is being visited
// for first time then this divisor
// must be the smallest prime number
// that divides this number
min_prime[j] = i;
}
// Update prime number till
// last prime number that divides this number
// The last prime number that
// divides this number will be maximum.
max_prime[j] = i;
j += i;
}
}
}
// Function to find the sum of the minimum
// and the maximum prime factors of every
// number from the given array
void findSum(int arr[], int n)
{
// Pre-calculation
sieve(MAX);
// For every element of the given array
for (int i = 0; i < n; i++) {
// The sum of its smallest
// and largest prime factor
int sum = min_prime[arr[i]]
+ max_prime[arr[i]];
cout << sum << " ";
}
}
// Driver code
int main()
{
int arr[] = { 5, 10, 15, 20, 25, 30 };
int n = sizeof(arr) / sizeof(int);
findSum(arr, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static int MAX = 100000;
// max_prime[i] represent maximum prime
// number that divides the number i
static int max_prime[] = new int[MAX + 1];
// min_prime[i] represent minimum prime
// number that divides the number i
static int min_prime[] = new int[MAX + 1];
// Function to store the minimum prime factor
// and the maximum prime factor in two arrays
static void sieve(int n)
{
for (int i = 2; i <= n; ++i)
{
// Check for prime number
// if min_prime[i] > 0,
// then it is not a prime number
if (min_prime[i] > 0)
{
continue;
}
// if i is a prime number
// min_prime number that divide prime number
// and max_prime number that divide prime number
// is the number itself.
min_prime[i] = i;
max_prime[i] = i;
int j = i + i;
while (j <= n)
{
if (min_prime[j] == 0)
{
// If this number is being visited
// for first time then this divisor
// must be the smallest prime number
// that divides this number
min_prime[j] = i;
}
// Update prime number till
// last prime number that divides this number
// The last prime number that
// divides this number will be maximum.
max_prime[j] = i;
j += i;
}
}
}
// Function to find the sum of the minimum
// and the maximum prime factors of every
// number from the given array
static void findSum(int arr[], int n)
{
// Pre-calculation
sieve(MAX);
// For every element of the given array
for (int i = 0; i < n; i++)
{
// The sum of its smallest
// and largest prime factor
int sum = min_prime[arr[i]]
+ max_prime[arr[i]];
System.out.print(sum + " ");
}
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 5, 10, 15, 20, 25, 30 };
int n = arr.length ;
findSum(arr, n);
}
}
// This code is contributed by AnkitRai01Python
# Python3 implementation of the approach
MAX = 100000
# max_prime[i] represent maximum prime
# number that divides the number i
max_prime = [0]*(MAX + 1)
# min_prime[i] represent minimum prime
# number that divides the number i
min_prime = [0]*(MAX + 1)
# Function to store the minimum prime factor
# and the maximum prime factor in two arrays
def sieve(n):
for i in range(2, n + 1):
# Check for prime number
# if min_prime[i]>0,
# then it is not a prime number
if (min_prime[i] > 0):
continue
# if i is a prime number
# min_prime number that divide prime number
# and max_prime number that divide prime number
# is the number itself.
min_prime[i] = i
max_prime[i] = i
j = i + i
while (j <= n):
if (min_prime[j] == 0):
# If this number is being visited
# for first time then this divisor
# must be the smallest prime number
# that divides this number
min_prime[j] = i
# Update prime number till
# last prime number that divides this number
# The last prime number that
# divides this number will be maximum.
max_prime[j] = i
j += i
# Function to find the sum of the minimum
# and the maximum prime factors of every
# number from the given array
def findSum(arr, n):
# Pre-calculation
sieve(MAX)
# For every element of the given array
for i in range(n):
# The sum of its smallest
# and largest prime factor
sum = min_prime[arr[i]] + max_prime[arr[i]]
print(sum, end = " ")
# Driver code
arr = [5, 10, 15, 20, 25, 30]
n = len(arr)
findSum(arr, n)
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 100000;
// max_prime[i] represent maximum prime
// number that divides the number i
static int []max_prime = new int[MAX + 1];
// min_prime[i] represent minimum prime
// number that divides the number i
static int []min_prime = new int[MAX + 1];
// Function to store the minimum prime factor
// and the maximum prime factor in two arrays
static void sieve(int n)
{
for (int i = 2; i <= n; ++i)
{
// Check for prime number
// if min_prime[i] > 0,
// then it is not a prime number
if (min_prime[i] > 0)
{
continue;
}
// if i is a prime number
// min_prime number that divide prime number
// and max_prime number that divide prime number
// is the number itself.
min_prime[i] = i;
max_prime[i] = i;
int j = i + i;
while (j <= n)
{
if (min_prime[j] == 0)
{
// If this number is being visited
// for first time then this divisor
// must be the smallest prime number
// that divides this number
min_prime[j] = i;
}
// Update prime number till
// last prime number that divides this number
// The last prime number that
// divides this number will be maximum.
max_prime[j] = i;
j += i;
}
}
}
// Function to find the sum of the minimum
// and the maximum prime factors of every
// number from the given array
static void findSum(int []arr, int n)
{
// Pre-calculation
sieve(MAX);
// For every element of the given array
for (int i = 0; i < n; i++)
{
// The sum of its smallest
// and largest prime factor
int sum = min_prime[arr[i]]
+ max_prime[arr[i]];
Console.Write(sum + " ");
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 5, 10, 15, 20, 25, 30 };
int n = arr.Length ;
findSum(arr, n);
}
}
// This code is contributed by 29AjayKumar输出：
10 7 8 7 10 7