给定一个整数k和一个整数数组arr ，任务是找到数组中k 个最小和k 个最大合数的和和乘积。
假设数组中至少有k 个合数。
例子：
Input: arr[] = {2, 5, 6, 8, 10, 11}, k = 2
Output: Sum of k-minimum composite numbers is 14
Sum of k-maximum composite numbers is 18
Product of k-minimum composite numbers is 48
Product of k-maximum composite numbers is 80
{6, 8, 10} are the only comsposite numbers from the array. {6, 8} are the 2 smallest and {8, 10} are the 2 largest among them.
Input: arr[] = {6, 4, 2, 12, 13, 5, 19, 10}, k = 3
Output: Sum of k-minimum composite numbers is 20
Sum of k-maximum composite numbers is 28
Product of k-minimum composite numbers is 240
Product of k-maximum composite numbers is 720
方法：
- 使用埃拉托色尼筛法生成一个布尔向量，其大小为数组中最大元素的大小，可用于检查数字是否为复合数。
- 还将0和1设置为素数，这样它们就不会被算作合数。
- 现在遍历数组并将所有合成的数字插入两个堆中，一个最小堆和一个最大堆。
- 现在，从最小堆中弹出前k 个元素，并取最小k 个合数的总和和乘积。
- 对最大堆做同样的事情，得到最大k 个合数的和和乘积。
- 最后，打印结果。
下面是上述方法的实现：
C++
// C++ program to find the sum and
// product of k smallest and k largest
// composite numbers in an array
#include
using namespace std;
vector SieveOfEratosthenes(int max_val)
{
// Create a boolean vector "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);
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;
}
}
return prime;
}
// Function that calculates the sum
// and product of k smallest and k
// largest composite numbers in an array
void compositeSumAndProduct(int arr[], int n, int k)
{
// 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
vector prime = SieveOfEratosthenes(max_val);
// Set 0 and 1 as primes so that
// they don't get counted as
// composite numbers
prime[0] = true;
prime[1] = true;
// Max Heap to store all the composite numbers
priority_queue maxHeap;
// Min Heap to store all the composite numbers
priority_queue, greater>
minHeap;
// Push all the composite numbers
// from the array to the heaps
for (int i = 0; i < n; i++)
if (!prime[arr[i]]) {
minHeap.push(arr[i]);
maxHeap.push(arr[i]);
}
long long int minProduct = 1
, maxProduct = 1
, minSum = 0
, maxSum = 0;
while (k--) {
// Calculate the products
minProduct *= minHeap.top();
maxProduct *= maxHeap.top();
// Calculate the sum
minSum += minHeap.top();
maxSum += maxHeap.top();
// Pop the current minimum element
minHeap.pop();
// Pop the current maximum element
maxHeap.pop();
}
cout << "Sum of k-minimum composite numbers is "
<< minSum << "\n";
cout << "Sum of k-maximum composite numbers is "
<< maxSum << "\n";
cout << "Product of k-minimum composite numbers is "
<< minProduct << "\n";
cout << "Product of k-maximum composite numbers is "
<< maxProduct;
}
// Driver code
int main()
{
int arr[] = { 4, 2, 12, 13, 5, 19 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 3;
compositeSumAndProduct(arr, n, k);
return 0;
} Java
// Java program to find the sum and
// product of k smallest and k largest
// composite numbers in an array
import java.util.*;
class GFG
{
static boolean[] SieveOfEratosThenes(int max_val)
{
// Create a boolean vector "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];
Arrays.fill(prime, true);
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed, then
// it is a prime
if (prime[p])
{
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
return prime;
}
// Function that calculates the sum
// and product of k smallest and k
// largest composite numbers in an array
static void compositeSumAndProduct(Integer[] arr,
int n, int k)
{
// Find maximum value in the array
int max_val = Collections.max(Arrays.asList(arr));
// Use sieve to find all prime numbers
// less than or equal to max_val
boolean[] prime = SieveOfEratosThenes(max_val);
// Set 0 and 1 as primes so that
// they don't get counted as
// composite numbers
prime[0] = true;
prime[1] = true;
// Max Heap to store all the composite numbers
PriorityQueue maxHeap =
new PriorityQueue((x, y) -> y - x);
// Min Heap to store all the composite numbers
PriorityQueue minHeap = new PriorityQueue<>();
// Push all the composite numbers
// from the array to the heaps
for (int i = 0; i < n; i++)
{
if (!prime[arr[i]])
{
minHeap.add(arr[i]);
maxHeap.add(arr[i]);
}
}
long minProduct = 1, maxProduct = 1,
minSum = 0, maxSum = 0;
Integer lastMin = 0, lastMax = 0;
while (k-- > 0)
{
if (minHeap.peek() != null ||
maxHeap.peek() != null)
{
// Calculate the products
minProduct *= minHeap.peek();
maxProduct *= maxHeap.peek();
// Calculate the sum
minSum += minHeap.peek();
maxSum += maxHeap.peek();
// Pop the current minimum element
lastMin = minHeap.poll();
// Pop the current maximum element
lastMax = maxHeap.poll();
}
else
{
// when maxHeap or minHeap is exhausted
// then this consition will run
minProduct *= lastMin;
maxProduct *= lastMax;
minSum += lastMin;
maxSum += lastMax;
}
}
System.out.println("Sum of k-minimum composite" +
" numbers is " + minSum);
System.out.println("Sum of k-maximum composite" +
" numbers is " + maxSum);
System.out.println("Product of k-minimum composite" +
" numbers is " + minProduct);
System.out.println("Product of k-maximum composite" +
" numbers is " + maxProduct);
}
// Driver Code
public static void main(String[] args)
{
Integer[] arr = { 4, 2, 12, 13, 5, 19 };
int n = arr.length;
int k = 3;
compositeSumAndProduct(arr, n, k);
}
}
// This code is contributed by
// sanjeev2552 输出：
Sum of k-minimum composite numbers is 28
Sum of k-maximum composite numbers is 20
Product of k-minimum composite numbers is 576
Product of k-maximum composite numbers is 192