Program to Count Even and Odd Numbers for a Given Array of Elements

#include <stdio.h>

int main() {
int n;

// Step 1: Read the size of the array
printf(“Enter the number of elements in the array: “);
if (scanf(“%d”, &n) != 1) {
printf(“Error reading input.\n”);
return 1;
}

int arr[n]; // Declare the array with user-defined size

// Step 2: Read the elements of the array
printf(“Enter %d elements:\n”, n);
for (int i = 0; i < n; i++) {
printf(“Element [%d]: “, i);
if (scanf(“%d”, &arr[i]) != 1) {
printf(“Error reading input.\n”);
return 1;
}
}

// Step 3: Count even and odd numbers
int evenCount = 0, oddCount = 0;
for (int i = 0; i < n; i++) {
if (arr[i] % 2 == 0) {
evenCount++; // Increment even counter
} else {
oddCount++; // Increment odd counter
}
}

// Step 4: Display the counts
printf(“\nNumber of even elements: %d\n”, evenCount);
printf(“Number of odd elements: %d\n”, oddCount);

return 0;
}

Explanation of the Program

1. Input Array Size and Elements:
   • The user inputs the size of the array and its elements.
2. Count Even and Odd Numbers:
   • A loop iterates through the array.
   • Each element is checked using the condition arr[i] % 2 == 0:
     • If true, it’s an even number, and the evenCount is incremented.
     • Otherwise, it’s an odd number, and the oddCount is incremented.
3. Display Results:
   • The program prints the total counts of even and odd numbers.

Sample Input/Output

Input:

Enter the number of elements in the array: 6
Enter 6 elements:
Element [0]: 12
Element [1]: 7
Element [2]: 9
Element [3]: 4
Element [4]: 10
Element [5]: 3

Output:

Number of even elements: 3
Number of odd elements: 3