给定一个正数数组，任务是计算非素数和素数之和之间的绝对差。
注意： 1既不是素数也不是非素数。
例子：
Input : 1 3 5 10 15 7
Output : 10
Explanation: Sum of non-primes = 25
Sum of primes = 15
Input : 3 4 6 7
Output : 0
天真的方法：一个简单的解决方案是遍历数组并检查每个元素是否为素数。如果数字是素数，则将其加到表示素数之和的和S2上，否则检查它是否不是1，然后将其加到非素数的总和中，比如说S1。遍历整个数组后，取两者之间的绝对差(S1-S2)。
时间复杂度：O(Nsqrt(N))
高效方法：使用Eratosthenes筛子生成所有素数，直到数组的最大元素，并将它们存储在哈希中。现在，遍历数组并检查哈希图中是否存在该数字。然后，将这些数字加到总和S2上，否则检查它是否不是1，然后将其加到总和S1上。
遍历整个数组后，显示两者之间的绝对差。
时间复杂度：O(Nlog(log(N))
C++
// C++ program to find the Absolute Difference
// between the Sum of Non-Prime numbers
// and Prime numbers of an Array
#include
using namespace std;
// Function to find the difference between
// the sum of non-primes and the
// sum of primes of an array.
int CalculateDifference(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;
}
}
// Store the sum of primes in S1 and
// the sum of non primes in S2
int S1 = 0, S2 = 0;
for (int i = 0; i < n; i++) {
if (prime[arr[i]]) {
// the number is prime
S1 += arr[i];
}
else if (arr[i] != 1) {
// the number is non-prime
S2 += arr[i];
}
}
// Return the absolute difference
return abs(S2 - S1);
}
int main()
{
// Get the array
int arr[] = { 1, 3, 5, 10, 15, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Find the absolute difference
cout << CalculateDifference(arr, n);
return 0;
} Java
// Java program to find the Absolute
// Difference between the Sum of
// Non-Prime numbers and Prime numbers
// of an Array
import java.util.*;
class GFG
{
// Function to find the difference
// between the sum of non-primes
// and the sum of primes of an array.
static int CalculateDifference(int arr[],
int n)
{
// Find maximum value in the array
int max_val = Integer.MIN_VALUE;
for(int i = 0; i < n; i++)
{
if(arr[i] > max_val)
max_val = arr[i];
}
// 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.
boolean []prime = new boolean[max_val + 1];
for(int i = 0; i <= max_val; i++)
prime[i] = 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;
}
}
// Store the sum of primes in
// S1 and the sum of non
// primes in S2
int S1 = 0, S2 = 0;
for (int i = 0; i < n; i++)
{
if (prime[arr[i]])
{
// the number is prime
S1 += arr[i];
}
else if (arr[i] != 1)
{
// the number is non-prime
S2 += arr[i];
}
}
// Return the absolute difference
return Math.abs(S2 - S1);
}
// Driver Code
public static void main(String []args)
{
// Get the array
int arr[] = { 1, 3, 5, 10, 15, 7 };
int n = arr.length;
// Find the absolute difference
System.out.println(CalculateDifference(arr, n));
}
}
// This code is contributed
// by ihritikPython3
# Python3 program to find the Absolute
# Difference between the Sum of Non-Prime
# numbers and Prime numbers of an Array
import sys
# Function to find the difference
# between the sum of non-primes
# and the sum of primes of an array.
def CalculateDifference(arr, n):
# Find maximum value in the array
max_val = -1
for i in range(0, n):
if(arr[i] > max_val):
max_val = arr[i]
# 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.
prime = [True for i in range(max_val + 1)]
# Remaining part of SIEVE
prime[0] = False
prime[1] = False
p = 2
while (p * p <= max_val):
# 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
p += 1
# Store the sum of primes in
# S1 and the sum of non primes
# in S2
S1 = 0
S2 = 0
for i in range (0, n):
if prime[arr[i]]:
# the number is prime
S1 += arr[i]
elif arr[i] != 1:
# the number is non-prime
S2 += arr[i]
# Return the absolute difference
return abs(S2 - S1)
# Driver code
# Get the array
arr = [ 1, 3, 5, 10, 15, 7 ]
n = len(arr)
# Find the absolute difference
print(CalculateDifference(arr, n))
# This code is contributed
# by ihritikC#
// C# program to find the Absolute
// Difference between the Sum of
// Non-Prime numbers and Prime
// numbers of an Array
using System;
class GFG
{
// Function to find the difference
// between the sum of non-primes
// and the sum of primes of an array.
static int CalculateDifference(int []arr,
int n)
{
// Find maximum value in the array
int max_val = int.MinValue;
for(int i = 0; i < n; i++)
{
if(arr[i] > max_val)
max_val = arr[i];
}
// 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.
bool []prime = new bool[max_val + 1];
for(int i = 0; i <= max_val; i++)
prime[i] = 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;
}
}
// Store the sum of primes in
// S1 and the sum of non primes
// in S2
int S1 = 0, S2 = 0;
for (int i = 0; i < n; i++)
{
if (prime[arr[i]])
{
// the number is prime
S1 += arr[i];
}
else if (arr[i] != 1)
{
// the number is non-prime
S2 += arr[i];
}
}
// Return the absolute difference
return Math.Abs(S2 - S1);
}
// Driver Code
public static void Main(string []args)
{
// Get the array
int []arr = { 1, 3, 5, 10, 15, 7 };
int n = arr.Length;
// Find the absolute difference
Console.WriteLine(CalculateDifference(arr, n));
}
}
// This code is contributed
// by ihritikPHP
$max_val)
$max_val = $arr[$i];
}
// 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.
$prime = array_fill(0, $max_val + 1, true);
// Remaining part of SIEVE
$prime[0] = false;
$prime[1] = false;
for ($p = 2; $p * $p <= $max_val; $p++)
{
// If prime[p] is not changed,
// then it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p * 2;
$i <= $max_val; $i += $p)
$prime[$i] = false;
}
}
// Store the sum of primes in
// S1 and the sum of non
// primes in S2
$S1 = 0;
$S2 = 0;
for ($i = 0; $i < $n; $i++)
{
if ($prime[$arr[$i]])
{
// the number is prime
$S1 += $arr[$i];
}
else if ($arr[$i] != 1)
{
// the number is non-prime
$S2 += $arr[$i];
}
}
// Return the absolute difference
return abs($S2 - $S1);
}
// Driver code
// Get the array
$arr = array( 1, 3, 5, 10, 15, 7 );
$n = sizeof($arr);
// Find the absolute difference
echo CalculateDifference($arr, $n);
// This code is contributed
// by ihritik
?>输出：
10