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### C program to calculate the sum of odd and even numbers

#### Problem Description

`This C program calculates the sum of odd and even numbers separately from 1 to n  (value of n is entered by the user). The for loop starts with 1 and continue up to the number entered by the user. To check whether the number in the current iteration is odd or even, the modulus operator is used.`

##### C program to calculate the sum of odd and even numbers - Source code
```

/*C program to find the sum of odd and even numbers from 1 to N */

#include <stdio.h>

int main()
{
int i, n;
int sum_odd = 0, sum_even = 0;

printf("Enter the number up to which you want to calculate sum\n");
scanf("%d", &n);
for (i = 1; i <= n; i++)
{
// to check whether number is odd or even
if (i % 2 == 0)
sum_even = sum_even + i;
else
sum_odd = sum_odd + i;
}
printf("Sum of all even numbers  = %d\n", sum_even);
printf("Sum of all odd numbers = %d\n", sum_odd);
}

```

#### Program Output

```Case 1:

Enter the number up to which you want to calculate sum
100
Sum of all even numbers  = 2550
Sum of all odd numbers = 2500

Case 2:

Enter the number up to which you want to calculate sum
25
Sum of all even numbers  = 156
Sum of all odd numbers = 169```

#### Program Explanation

```1. Two integer variables i and n are declared. Two integer variables sum_odd and sum_even are declared and initialized to 0.

2. For loop starts the first iteration with 1 and continue up to the number entered by the user (n).

3. To check whether the number in the current iteration is even or odd, modulus operator (%) is used. This operator divides the number by 2 and returns the remainder. If reminder returned is 0, the number is even otherwise it is odd.

4. The value of sum_even and sum_odd operators is updated inside the loop and printed when the loop ends.```