Question

Даю 40 балів!!! Visual Studio 2019Мова програмування С++1. Задано натуральне число n і дійсні числа b1,b2,...bn. Визначити суму тих дійсних чисел які більше p.​2. Дано натуральне число n. отримати послідовність a1,a2,...an , яка утворюється згiдно з законом: a0 = 1 ; ak=ka(k-1) +1/k ( k =1,2...)3. Задано натуральнi числа х i у. Обчислити добуток х*у, використовуючи лише операцію складання. Завдання вирішити двома способами.​

Answer 1

Завдання 1:

#include <iostream>

using namespace std;

int Task1()

{

   int n, p;

   cout << "Введіть кількість елементів масиву: ";

   cin >> n;

   cout << "Введіть число p: ";

   cin >> p;

   int *arr = new int[n];

   for (int i = 0; i < n; i++)

   {

       cout << "Введіть " << i + 1 << " елемент масиву: ";

       cin >> arr[i];

   }

   int sum = 0;

   for (int i = 0; i < n; i++)

   {

       if (arr[i] > p)

       {

           sum += arr[i];

       }

   }

   cout << "Сума елементів масиву, які більші за p: " << sum << endl;

   delete[] arr;

   return 0;

}

Завдання 2:

#include <iostream>

using namespace std;

int Task2()

{

   int n;

   cout << "Введіть кількість елементів масиву: ";

   cin >> n;

   double *arr = new double[n];

   arr[0] = 1;

   for (int i = 1; i < n; i++)

   {

       arr[i] = i * arr[i - 1] + 1 / i;

   }

   for (int i = 0; i < n; i++)

   {

       cout << "a" << i + 1 << ": " << arr[i] << endl;

   }

   delete[] arr;

   return 0;

}

Завдання 3:

#include <iostream>

using namespace std;

int Task3()

{

   int x, y;

   cout << "Введіть число x: ";

   cin >> x;

   cout << "Введіть число y: ";

   cin >> y;

   int sum = 0;

   for (int i = 0; i < y; i++)

   {

       sum += x;

   }

   cout << "Сума: " << sum << endl;

   return 0;

}

Answer 2

Task 1: Finding the sum of real numbers greater than p

To find the sum of real numbers greater than p, given a natural number n and real numbers b1, b2, ..., bn, you can follow these steps:

1. Initialize a variable `sum` to 0. 2. Iterate over each real number `b` in the given list of numbers. 3. Check if `b` is greater than p. 4. If `b` is greater than p, add `b` to the `sum`. 5. After iterating over all the numbers, the `sum` will contain the sum of real numbers greater than p.

Here's an example implementation in C++:

#include 
#include

int main() {
    int n;
    double p;
    std::vector numbers;

// Read the values of n, p, and the list of numbers
    std::cout << "Enter the value of n: ";
    std::cin >> n;

std::cout << "Enter the value of p: ";
    std::cin >> p;

std::cout << "Enter " << n << " real numbers: ";
    for (int i = 0; i < n; i++) {
        double num;
        std::cin >> num;
        numbers.push_back(num);
    }

// Calculate the sum of real numbers greater than p
    double sum = 0;
    for (double b : numbers) {
        if (b > p) {
            sum += b;
        }
    }

// Output the result
    std::cout << "The sum of real numbers greater than " << p << " is: " << sum << std::endl;

return 0;
}

Please note that this is just an example implementation. You can modify it according to your specific requirements.

Task 2: Generating a sequence according to the given law

To generate a sequence `a1, a2, ..., an` according to the law `a0 = 1; ak = ka(k-1) + 1/k (k = 1, 2...)`, you can follow these steps:

1. Initialize a variable `a` to 1. 2. Initialize an empty list `sequence`. 3. Append `a` to the `sequence`. 4. Iterate from `k = 1` to `n`. 5. Calculate the value of `a` using the given law: `a = k * a * (a - 1) + 1 / k`. 6. Append the calculated `a` to the `sequence`. 7. After the iteration, the `sequence` will contain the desired sequence `a1, a2, ..., an`.

Here's an example implementation in C++:

#include 
#include

int main() {
    int n;
    std::vector sequence;

// Read the value of n
    std::cout << "Enter the value of n: ";
    std::cin >> n;

// Generate the sequence
    double a = 1;
    sequence.push_back(a);
    for (int k = 1; k <= n; k++) {
        a = k * a * (a - 1) + 1.0 / k;
        sequence.push_back(a);
    }

// Output the sequence
    std::cout << "The sequence is: ";
    for (double num : sequence) {
        std::cout << num << " ";
    }
    std::cout << std::endl;

return 0;
}

Again, please note that this is just an example implementation. You can modify it according to your specific requirements.

Task 3: Computing the product of x and y using only addition

To compute the product of two natural numbers x and y using only the addition operation, you can follow these steps:

1. Initialize a variable `product` to 0. 2. Iterate `y` times. 3. Add `x` to the `product` in each iteration. 4. After the iteration, the `product` will contain the desired product of x and y.

Here's an example implementation in C++:

#include

int main() {
    int x, y;

// Read the values of x and y
    std::cout << "Enter the value of x: ";
    std::cin >> x;

std::cout << "Enter the value of y: ";
    std::cin >> y;

// Compute the product of x and y using only addition
    int product = 0;
    for (int i = 0; i < y; i++) {
        product += x;
    }

// Output the result
    std::cout << "The product of " << x << " and " << y << " is: " << product << std::endl;

return 0;
}

Once again, please note that this is just an example implementation. You can modify it according to your specific requirements.

I hope this helps! Let me know if you have any further questions.