URI - BEECROWD - BEE 1070 - Six Odd Numbers Solution in C,C++,Python

URI - BEECROWD - BEE 1070 - Six Odd Numbers Solution in C,C++,Python | URI Online Judge Solution 1070 - Six Odd Numbers

Read an integer value X and print the 6 consecutive odd numbers from X, a value per line, including X if it is the case.

Input

The input will be a positive integer value.

Output

The output will be a sequence of six odd numbers.

Input Sample	Output Sample
8	9 11 13 15 17 19

URI 1070 - Six Odd Numbers Solution in C,C++,Python | URI Online Judge Solution 1070 - Six Odd Numbers :

                                        Demonstration:

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

Here we apply a range-based while loop which will run 6 times and checking the condition for six 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 1070 Solution in C :   

URI - BEECROWD - BEE Online Judge 1070 Solve  in C :

#include <stdio.h>

int main()
{
    int n, i=0;
    scanf("%d", &n);
    while(i < 6){
        if(n%2!=0){
            printf("%d\n", n);
            i++;
        }   
        n++;
     }   
    return 0;
}

 URI-BEECROWD-BEE Problem 1070 Solution in C ++:

URI - BEECROWD - BEE Online Judge 1070 Solve  in C++ :

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

int main()
{
    int n, i=0;
    cin>>n;
    while(i < 6){
        if(n%2!=0){
            cout<<n<<endl;
            i++;
        }   
        n++;
     }   
    return 0;
}

 URI-BEECROWD-BEE Problem 1070 Solution in Python:

URI - BEECROWD - BEE Online Judge 1070 Solve  in Python  :

n=int(input())
i=0
while(i<6):
    if(n%2!=0):
        print(n)
        i=i+1
    n=n+1

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