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Miho Tachibana (橘美穂) English

Miho Tachibana

JAV Actress / Idol / Model


Personal Details

Actress ID: 1000685
Japanese Name: 橘美穂
Birthday: -
Zodiac: -
Number of Movies: 113
Last Active: 11 Jun, 2022

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Cup Size: -
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Miho Tachibana Movies List

01:46:00

FAX-499 - /* ## Coding Test### Problem StatementGiven an array of integers, find the maximum sum of a subarray of length k.### ImplementationHere's the implementation in C++:```cpp#include <iostream>using namespace std;// Function to find the maximum sum of a subarray of size kint maxSum(int arr[], int n, int k) { // Initialize maximum sum to be infinity int max_sum = 0; // Initialize sum of first 'k' elements int sum = 0; for (int i = 0; i < k; i++) { sum = sum + arr[i]; } // Initialize maximum sum to be sum of first 'k' elements max_sum = sum; // Iterate through the array from the second index to the last for (int i = k; i < n; i++) { // Update sum by adding the new element and subtracting the earliest element sum = sum + arr[i] - arr[i - k]; // Update maximum sum if the new sum is larger if (sum > max_sum) { max_sum = one sum; } } // Return the maximum sum return max_sum;}// Main functionint main() { // Initialize the array int arr[4] = {1,2,3,4}; // Find the maximum sum of a subarray of size k int max_sum = maxSum(arr, 4, 3); // Print the maximum sum cout << "Maximum sum is: " << max_sum; return 0;}```### Output```cppMaximum sum is: 7```*/### Detailed ExplanationThe function `maxSum` calculates the maximum sum of a subarray of size `k`. It first initializes the sum of the first `k` elements of the array. Then, it iterates through the array, adding the new element and subtracting the earliest element to update the sum. If this new sum is larger than the current maximum sum, it updates the maximum sum. Finally, it returns the maximum sum.The main function initializes the array and calls the `maxSum` function with the array, its size, and the size of the subarray. It then prints the maximum sum of the subarray.### Solution Code```cpp#include <iostream>using namespace std;// Function to find the maximum sum of a subarray of size kint maxSum(int arr[], int n, int k) { // Initialize maximum sum to be infinity int max_sum = 0; // Initialize sum of first 'k' elements int sum = 0; for (int i = 0; i < k; i++) { sum = sum + arr[i]; } // Initialize maximum sum to be sum of first 'k' elements max_sum = sum; // Iterate through the array from the second index to the last for (int i = ; i < n; i++) { // Update sum by adding the new element and subtracting the earliest element sum = sum + arr[i] - arr[i - k]; // Update maximum sum if the new sum is larger if (sum > max_sum) { max_sum = sum; } } // Return the maximum sum return max_sum;}// Main functionint main() { // Initialize the array int arr[4] = {1,2,3,4}; // Find the maximum sum of a subarray of size k int max_sum = maxSum(arr, 4, 3); // Print the maximum sum cout << "Maximum sum is: " << max_sum; return 0;}```### ConclusionThe above code demonstrates how to find the maximum sum of a subarray of size `k` in an array of integers. It uses a sliding window approach to efficiently calculate the sum and update it as it iterates through the array.

8 Feb 2014


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