URI - BEECROWD - BEE 1067 - Odd Numbers Solution in C,C++,Python

 URI - BEECROWD - BEE 1067 - Odd Numbers Solution in C,C++,Python |  URI - BEECROWD - BEE Online Judge Solution 1067 - Odd Numbers

Read an integer value X (1 <= X <= 1000).  Then print the odd numbers from 1 to X, each one in a line, including X if is the case.

Input

The input will be an integer value.

Output

Print all odd values between 1 and X, including X if is the case.

Input Sample	Output Sample
8	1 3 5 7

 URI - BEECROWD - BEE 1067 - Odd Numbers Solution in C,C++,Python | URI - BEECROWD - BEE Online Judge Solution 1067 - Odd Numbers :

                                        Demonstration:

Any integer that cannot be divided exactly by 2 is an odd number.

Here we apply a range-based for loop which provides all the integers available in the input interval.

After this, checking the condition for odd numbers are applied and printing these values. 

N.B: Don't copy paste the code as same. Just try to understand it and try yourself.

   URI - BEECROWD - BEE Problem 1067 Solution in C :   

URI - BEECROWD - BEE Online Judge 1067 Solve  in C :

#include<stdio.h>
int main()
{
    int n,i;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        if(i%2!=0){
            printf("%d\n",i);
        }
    }
}

 URI - BEECROWD - BEE Problem 1067 Solution in C ++:

URI - BEECROWD - BEE Online Judge 1067 Solve  in C++ :

#include<bits/stdc++.h>
using namespace std;

int main(){
    int n,i;
    cin>>n;
    for(i=1;i<=n;i++){
        if(i%2==1){
            cout<<i<<endl;
        }

    }
    return 0;
}

 URI-BEECROWD-BEE Problem 1067 Solution in Python:

URI - BEECROWD - BEE Online Judge 1067 Solve  in Python  :

n = int(input())
for i in range(n+1):
    if(i%2==1):
        print(i)

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