Odd elements (#7841)

Algorithm Analysis

We will dynamically read the elements of the sequence using a loop and output the odd ones. The number $v a l$ is odd if $v a l %2 \neq = 0$ . The condition $v a l %2 == 1$ for checking the oddness of a number is not correct for a negative $v a l$ . In C, when dividing an odd negative number by 2, the result will be -1.

Algorithm Implementation – Loop

Read the number of input numbers.

scanf("%d",&n);

In the loop, process $n$ numbers.

for(i = 0; i < n; i++)
{
    scanf("%d",&val);
    //if (val % 2 == 1 || val % 2 == -1) printf("%d ",val);
    if (val % 2 != 0) printf("%d ",val);
}

Algorithm Implementation – Array

Declare a working array.

int m[101];

Read the input sequence of integers into the array m.

scanf("%d",&n);
for (i = 0; i < n; i++)
    scanf("%d", &m[i]);

Iterate through the numbers in the array. If m[i] is odd, then output it.

for (i = 0; i < n; i++)
    if (m[i] % 2 != 0) printf("%d ", m[i]);
printf("\n");

Java Implementation

import java.util.*;

class Main
{
    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        int n = con.nextInt();
        for(int i = 0; i < n; i++)
        {
            int val = con.nextInt();
            if (val % 2 != 0) System.out.print(val + " ");
        }
        System.out.println();
        con.close();
    }
}

Python Implementation

Read the input sequence of integers into the list lst.

n = int(input())
lst = list(map(int,input().split()))

In the loop, process $n$ numbers of the list lst.

for x in lst:
    if x % 2 != 0:
        print(x, end = " ")

Python Implementation – List

Read the input sequence of integers into the list lst.

n = int(input())
lst = list(map(int,input().split()))

In the loop, process $n$ numbers of the list lst.

for i in range(n):
    if lst[i] % 2 != 0:
        print(lst[i], end = " ")