大小为K的所有子数组的比率

给定一个数组arr []和一个整数K ，任务是计算大小为K的所有子数组的比率。

例子：

Input: arr[] = {24, 3, 2, 1}, K = 3 
Output: 4 1.5 
Explanation: 
All subarrays of size K and their ratio: 
Subarray 1: {24, 3, 2} = 24 / 3 / 2 = 4 
Subarray 2: {3, 2, 1} = 3 / 2 / 1= 1.5

Input: arr[] = {1, -2, 3, -4, 5, 6}, K = 2 
Output: -0.5 -0.666667 -0.75 -0.8 0.833333 

方法：想法是遍历给定数组中存在的每个大小为K的子数组，并按照以下步骤解决问题：

1. 用子数组的第一个元素初始化一个变量，比如说比率。
2. 迭代的其余子阵列，并保持由上所遇到的元件逐个分频比。
3. 最后，打印该子数组的ratio的最终值。
4. 对所有子数组重复上述步骤。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to find the ratio of
// all subarrays of size K
int calcRatio(double arr[], int n, int k)
{
 
    // Traverse every subarray of size K
    for (int i = 0; i <= n - k; i++) {
 
        // Initialize ratio
        double ratio = arr[i];
 
        // Calculate ratio of the
        // current subarray
        for (int j = i + 1; j < k + i; j++)
            ratio /= arr[j];
 
        // Print ratio of the subarray
        cout << ratio << " ";
    }
}
 
// Driver Code
int main()
{
    // Given array
    double arr[] = { 24, 3, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    // Function Call
    calcRatio(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
class GFG{
   
// Function to find the ratio of
// all subarrays of size K
static void calcRatio(double arr[], int n,
                                    int k)
{
     
    // Traverse every subarray of size K
    for(int i = 0; i <= n - k; i++)
    {
   
        // Initialize ratio
        double ratio = arr[i];
   
        // Calculate ratio of the
        // current subarray
        for(int j = i + 1; j < k + i; j++)
            ratio /= arr[j];
   
        // Print ratio of the subarray
        System.out.print(ratio + " ");
    }
}
  
// Driver code
public static void main (String[] args)
{
     
    // Given array
    double arr[] = { 24, 3, 2, 1 };
    int n = arr.length;
    int k = 3;
     
    // Function call
    calcRatio(arr, n, k);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation
# of the above approach
 
# Function to find the ratio of
# all subarrays of size K
def calcRatio(arr, n, k):
 
    # Traverse every subarray
    # of size K
    for i in range(n - k + 1):
 
        # Initialize ratio
        ratio = arr[i]
 
        # Calculate ratio of the
        # current subarray
        for j in range(i + 1, k + i):
            ratio = ratio / arr[j]
 
        # Print ratio of the subarray
        print(ratio, end = " ")
 
# Given array
arr = [24, 3, 2, 1]
n = len(arr)
k = 3
 
# Function Call
calcRatio(arr, n, k)
 
# This code is contributed by divyeshrabadiya07

C#

// C# implementation of the above approach
using System;
  
class GFG{
    
// Function to find the ratio of
// all subarrays of size K
static void calcRatio(double []arr, int n,
                                    int k)
{
   
    // Traverse every subarray of size K
    for(int i = 0; i <= n - k; i++)
    {
    
        // Initialize ratio
        double ratio = arr[i];
    
        // Calculate ratio of the
        // current subarray
        for(int j = i + 1; j < k + i; j++)
            ratio /= arr[j];
    
        // Print ratio of the subarray
        Console.Write(ratio + " ");
    }
}
   
// Driver code
public static void Main(string[] args)
{
      
    // Given array
    double []arr = { 24, 3, 2, 1 };
    int n = arr.Length;
    int k = 3;
      
    // Function call
    calcRatio(arr, n, k);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript implementation of the above approach
 
// Function to find the ratio of
// all subarrays of size K
function calcRatio(arr, n, k)
{
 
    // Traverse every subarray of size K
    for (var i = 0; i <= n - k; i++) {
 
        // Initialize ratio
        var ratio = arr[i];
 
        // Calculate ratio of the
        // current subarray
        for (var j = i + 1; j < k + i; j++)
            ratio /= arr[j];
 
        // Print ratio of the subarray
        document.write(ratio + " ");
    }
}
 
// Driver Code
// Given array
var arr = [ 24, 3, 2, 1 ];
var n = arr.length;
var k = 3;
// Function Call
calcRatio(arr, n, k);
 

输出

4 1.5 

时间复杂度： O(N ² )
辅助空间： O(1)