The provided C program performs the following functionality: ### Overview: The program...

August 30, 2025 at 09:58 AM

#include <stdio.h>
void input(int *buffer, int *length);
void output(int *buffer, int length);
void output_number(int num);  // Новая функция для вывода чисел без %
int sum_numbers(int *buffer, int length);
int find_numbers(int* buffer, int length, int number, int* numbers);
int main()
{
    int length;
    int buffer[10];
    int numbers[10];
    
    input(buffer, &length);
    
    if (length <= 0 || length > 10) {
        printf("n/a");
        return 0;
    }
    
    int sum = sum_numbers(buffer, length);
    
    if (sum == 0) {
        printf("n/a");
        return 0;
    }
    
    int count = find_numbers(buffer, length, sum, numbers);
    
    if (count == 0) {
        printf("n/a");
        return 0;
    }
    
    // Вывод суммы без использования %
    output_number(sum);
    putchar('\n');
    output(numbers, count);
    
    return 0;
}

void input(int *buffer, int *length)
{
    scanf("%d", length);
    
    for (int i = 0; i < *length; i++) {
        scanf("%d", &buffer[i]);
    }
}

void output_number(int num) {
    if (num == 0) {
        putchar('0');
        return;
    }
    
    if (num < 0) {
        putchar('-');
        num = -num;
    }
    
    // Определяем количество цифр
    int temp = num;
    int digits = 0;
    while (temp > 0) {
        digits++;
        temp /= 10;
    }
    
    // Выводим цифры
    char num_str[12];
    temp = num;
    for (int i = digits - 1; i >= 0; i--) {
        num_str[i] = '0' + (temp % 10);
        temp /= 10;
    }
    for (int i = 0; i < digits; i++) {
        putchar(num_str[i]);
    }
}

void output(int *buffer, int length)
{
    for (int i = 0; i < length; i++) {
        output_number(buffer[i]);
        
        if (i < length - 1) {
            putchar(' ');
        }
    }
}
int sum_numbers(int *buffer, int length)
{
    int sum = 0;
    int found_even = 0;
    
    for (int i = 0; i < length; i++) {
        if (buffer[i] % 2 == 0 && buffer[i] != 0) {
            sum += buffer[i];
            found_even = 1;
        }
    }
    
    return found_even ? sum : 0;
}


int find_numbers(int* buffer, int length, int number, int* numbers)
{
    int count = 0;
    
    for (int i = 0; i < length; i++) {
        if (number != 0 && buffer[i] != 0 && number % buffer[i] == 0) {
            numbers[count] = buffer[i];
            count++;
        }
    }
    
    return count;
}

The provided C program performs the following functionality:

Overview:

The program processes an array of integers provided by the user, calculates the sum of the even (non-zero) numbers in the array, and identifies elements in the array that are divisors of this sum. It then outputs the results in a formatted way. The code handles various error cases, such as invalid input or cases where there are no valid numbers meeting the specified criteria.

Detailed Steps:

Main Function:
- Reads an integer array (buffer) with a maximum size of 10 elements using the input function.
- Performs input validation:
  - Ensures the length of the array (length) is between 1 and 10. If not, it outputs "n/a" and exits.
- Computes the sum of all non-zero, even integers in the array using sum_numbers.
  - If the sum is 0, it outputs "n/a" and exits.
- Finds numbers in the array that divide this sum evenly using find_numbers.
  - If no divisors are found, it outputs "n/a" and exits.
- Outputs the sum without using printf formatting or the % operator, via the output_number function.
- Outputs the list of divisors (numbers that divide the sum evenly) using the output function.

Helper Functions:
- input Function:
  Reads the size of the array (length) and then the integer elements into the buffer array.
- sum_numbers Function:
  Calculates and returns the sum of all non-zero, even numbers in the array. If no valid even numbers are found, it returns 0.
- find_numbers Function:
  Finds elements in the array (buffer) that divide a given number evenly. These divisors are stored in the numbers array, and the total count of such divisors is returned.
- output Function:
  Outputs the elements of a given integer array (buffer) of size length, separating them with spaces.
- output_number Function:
  Outputs a single integer (num) one character at a time without using printf format specifiers or division/remainder (%) operators for formatting. It supports both positive and negative numbers.

Error Cases:

If the user enters an invalid array size (length <= 0 or length > 10).
If sum_numbers returns 0, meaning there are no valid (non-zero, even) numbers in the array.
If find_numbers finds no divisors of the computed sum.

In all these cases, the program outputs "n/a" and exits.

Example Execution:

Input:

5
1 2 3 4 6

Here:

Array length = 5
Array: [1, 2, 3, 4, 6]

Execution Steps:

Sum of even numbers:
Non-zero even numbers are [2, 4, 6]. Their sum is 2 + 4 + 6 = 12.
Find divisors of sum (12):
Divisors of 12 in the array are [2, 3, 4, 6].
Output:
Prints 12 (the sum) and the divisors 2 3 4 6.

Output:

12
2 3 4 6

Input with Errors:

Input:

Output:

n/a

(Length is invalid.)

Input:

5
1 1 1 1 1

Output:

n/a

(No valid even numbers, so the sum is 0.)

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