给定一个正数数组，任务是找到非质数之和与质数之和的按位与。请注意， 1既不是质数也不是合数。
例子：
Input: arr[] = {1, 3, 5, 10, 15, 7}
Output: 9
Sum of non-primes = 10 + 15 = 25
Sum of primes = 3 + 5 + 7 = 15
25 & 15 = 9
Input: arr[] = {3, 4, 6, 7}
Output: 10
天真的方法：一个简单的解决方案是遍历数组并不断检查每个元素是否为素数。如果数字是素数，则将其添加到存储数组中素数之和的 S1 中，否则将其添加到存储非素数之和的 S2 中。最后，打印 S1 & S2。
时间复杂度： O(N * sqrt(N))
有效的方法：使用 Eratosthenes 筛法生成直到数组最大元素的所有素数，并将它们存储在散列中。现在，遍历数组并检查数字是否为素数。最后，计算并打印素数之和与合数之和的按位与。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
int calculateAND(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 not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
int main()
{
int arr[] = { 3, 4, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << calculateAND(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
class GFG
{
static int getMax(int[] A)
{
int max = Integer.MIN_VALUE;
for (int i: A)
{
max = Math.max(max, i);
}
return max;
}
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
static int calculateAND(int arr[], int n)
{
// using Collections.max() to find
// maximum element using only 1 line.
// Find maximum value in the array
int max_val = getMax(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.
boolean prime[] = new boolean [max_val + 1];
int i;
for (i = 0; i < max_val + 1; 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 ( 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 (i = 0; i < n; i++)
{
if (prime[arr[i]])
{
// The number is prime
S1 += arr[i];
}
else if (arr[i] != 1)
{
// The number is not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 3, 4, 6, 7 };
int n = arr.length;
System.out.println(calculateAND(arr, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
# Function to return the bitwise AND of the
# sum of primes and the sum of non-primes
def calculateAND(arr, n):
# Find maximum value in the array
max_val = max(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.
prime = [True for i in range(max_val + 1)]
# Remaining part of SIEVE
prime[0] = False
prime[1] = False
for p in range(2, max_val + 1):
if p * p >= max_val:
break
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# Update all multiples of p
for i in range(2 * p, max_val + 1, 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 in range(n):
if (prime[arr[i]]):
# The number is prime
S1 += arr[i]
elif (arr[i] != 1):
# The number is not prime
S2 += arr[i]
# Return the bitwise AND of the sums
return (S1 & S2)
# Driver code
arr = [3, 4, 6, 7]
n = len(arr)
print(calculateAND(arr, n))
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int getMax(int[] A)
{
int max = int.MinValue;
foreach (int i in A)
{
max = Math.Max(max, i);
}
return max;
}
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
static int calculateAND(int []arr, int n)
{
// using Collections.max() to find
// maximum element using only 1 line.
// Find maximum value in the array
int max_val = getMax(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.
bool []prime = new bool [max_val + 1];
int i;
for (i = 0; i < max_val + 1; 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 (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 (i = 0; i < n; i++)
{
if (prime[arr[i]])
{
// The number is prime
S1 += arr[i];
}
else if (arr[i] != 1)
{
// The number is not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 3, 4, 6, 7 };
int n = arr.Length;
Console.WriteLine(calculateAND(arr, n));
}
}
// This code is contributed by Rajput-JiJavascript
输出：
10时间复杂度： O(N * log(log(N))
空间复杂度： O(max_val) 其中 max_val 是给定数组中元素的最大值。
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