数组中 K 个最小和最大斐波那契数的总和和乘积

给定一个整数K和一个包含N 个整数的数组 arr[] ，任务是找到数组中 K 个最小和K 个最大斐波那契数的和和乘积。

注意：假设数组中至少有K 个斐波那契数。

例子：

Input: arr[] = {2, 5, 6, 8, 10, 11}, K = 2
Output:
Sum of K-minimum fibonacci numbers is 7
Product of K-minimum fibonacci numbers is 10
Sum of K-maximum fibonacci numbers is 13
Product of K-maximum fibonacci numbers is 40
Explanation :
{2, 5, 8} are the only fibonacci numbers from the array.
{2, 5} are the 2 smallest and {5, 8} are the 2 largest among them.

Input: arr[] = {3, 2, 12, 13, 5, 19}, K = 3
Output:
Sum of K-minimum fibonacci numbers is 10
Product of K-minimum fibonacci numbers is 30
Sum of K-maximum fibonacci numbers is 21
Product of K-maximum fibonacci numbers is 195

方法：这个想法是使用散列来预先计算和存储斐波那契节点直到最大值，在一个集合中，使检查变得容易和高效(在 O(1) 时间内)。

1. 遍历整个数组，获取列表中的最大值。
2. 现在，构建一个哈希表，其中包含小于或等于数组最大值的所有斐波那契节点。

执行上述预计算后，遍历数组并将所有斐波那契数插入两个堆，一个最小堆和一个最大堆。

现在，从最小堆和最大堆中弹出顶部 K 个元素来计算 K 个斐波那契数的和和乘积。

下面是上述方法的实现：

// C++ program to find the sum and
// product of K smallest and K
// largest Fibonacci numbers in an array
  
#include 
using namespace std;
  
// Function to create the hash table
// to check Fibonacci numbers
void createHash(set& hash,
                int maxElement)
{
    // Inserting the first two elements
    // into the hash
    int prev = 0, curr = 1;
    hash.insert(prev);
    hash.insert(curr);
  
    // Computing the remaining
    // elements using
    // the previous two elements
    while (curr <= maxElement) {
        int temp = curr + prev;
        hash.insert(temp);
        prev = curr;
        curr = temp;
    }
}
  
// Function that calculates the sum
// and the product of K smallest and
// K largest Fibonacci numbers in an array
void fibSumAndProduct(int arr[],
                      int n, int k)
{
    // Find the maximum value in the array
    int max_val = *max_element(arr, arr + n);
  
    // Creating a hash containing
    // all the Fibonacci numbers
    // upto the maximum data value
    // in the array
    set hash;
    createHash(hash, max_val);
  
    // Max Heap to store all the
    // Fibonacci numbers
    priority_queue maxHeap;
  
    // Min Heap to store all the
    // Fibonacci numbers
    priority_queue,
                   greater >
        minHeap;
  
    // Push all the fibonacci numbers
    // from the array to the heaps
    for (int i = 0; i < n; i++)
        if (hash.find(arr[i])
            != hash.end()) {
  
            minHeap.push(arr[i]);
            maxHeap.push(arr[i]);
        }
  
    long long int minProduct = 1,
                  maxProduct = 1,
                  minSum = 0,
                  maxSum = 0;
  
    // Finding the K minimum
    // and the K maximum
    // elements from the heaps
    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 "
         << "fibonacci numbers is "
         << minSum << "\n";
    cout << "Product of K-minimum "
         << "fibonacci numbers is "
         << minProduct << "\n";
    cout << "Sum of K-maximum "
         << "fibonacci numbers is "
         << maxSum << "\n";
    cout << "Product of K-maximum "
         << "fibonacci numbers is "
         << maxProduct;
}
  
// Driver code
int main()
{
  
    int arr[] = { 2, 5, 6, 8, 10, 11 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    int K = 2;
  
    fibSumAndProduct(arr, N, K);
  
    return 0;
}

输出：

Sum of K-minimum fibonacci numbers is 7
Product of K-minimum fibonacci numbers is 10
Sum of K-maximum fibonacci numbers is 13
Product of K-maximum fibonacci numbers is 40

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