Question

#include using namespace std; bool akk(const int n) sum=0; { if (n == 1) return false; for (int i=2; i<=n/2; ++i) if (n % i == 0) return false; return true; } int main() { for (int i=1; i<=100; ++i) if (akk(i)) cout << i << endl; return 0; } программа выводит простые числа. Помогите их просуммировать

Answer 1

Ответ:

Объяснение:

#include

using namespace std;

bool akk(const int n)

sum=0;

{

if (n == 1)

return false;

for (int i=2; i<=n/2; ++i)

if (n % i == 0)

return false;

return true;

}

int main()

{

for (int i=1; i<=100; ++i)

if (akk(i))

sum+=i; // вот эта строчка суммирует их, если хочешь выведи

cout << i << endl;

return 0;

}

Answer 2

Certainly! To modify the given code to calculate the sum of prime numbers, you can add a variable to keep track of the sum and update it accordingly. Here's the modified code:

cppCopy code
#include <iostream>

using namespace std;

bool isPrime(const int n) {
    if (n == 1)
        return false;

    for (int i = 2; i <= n / 2; ++i) {
        if (n % i == 0)
            return false;
    }

    return true;
}

int main() {
    int sum = 0;

    for (int i = 1; i <= 100; ++i) {
        if (isPrime(i)) {
            cout << i << endl;
            sum += i;  // Update the sum
        }
    }

    cout << "Sum of prime numbers: " << sum << endl;

    return 0;
}

In this modified code, I added the sum variable to store the sum of prime numbers. Inside the for loop, when a prime number is found, it is printed, and its value is added to the sum variable. Finally, outside the loop, the sum is displayed.

Now, when you run this code, it will output the prime numbers from 1 to 100 and display the sum of those prime numbers.