Here's to all the new programmers! I, too, am quite new to programming and want to help you guys be better at programming. I will post programming tasks from time to time and ask you guys to try and code something that can solve this. If you manage to, congratulations! If you don't, I will post the answer (mostly in C++ cause that's my dominant language) after a few days so you can analyze it.

For this task:

Plarry the Dinosaur needs to write a composition of \(n\) lines. Being an evil teacher, even after Plarry has written his composition, now the teacher wants Plarry to output every third word! Being a lazy dinosaur, he wants you to help him. Make a code that given, \(n\) and Plarry's composition, will output every \(3\)rd word.

Sample input:

\(7\)

This

Seven

Words

Than

Sample Output:

Seven

Congratz to those who have solved it, i'll post the answer 2 days later ^.^

Feel free to look at the answers below for reference, mine is quite similar to Daniel's but I feel that taking a look at the other languages would have benefits too!

Comments

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In C:

#include <stdio.h>

int main() {
    int N;
    scanf("%d\n", &N);
    for(int i = 1; i <= N; i++) {
        char *str = malloc(100 * sizeof(str));
        scanf("%s\n", str);
        if(i % 3 == 0) printf("%s\n", str);
        // to avoid memory leak
        free(str);
   }
   return 0;
}

– Ruhan Habib · 3 years, 1 month ago

Here is a python2 solution below:

num = input() # number of words 
printable = []
for i in xrange(1, num+1):
    if i%3 == 0:
        printable.append(raw_input()) # Only inserting the needed words that we'll print
    else:
        raw_input() # simply rejecting this input
for word in printable:
    print word

– Iamsudip Iamsudip · 3 years, 1 month ago

C++

#include <iostream>
#include <string>
using namespace std;

int main() {

    int TC; cin >> TC;
    for(int i = 1; i <= TC; i++){
        string str; cin >> str;
        if(i%3==0)
        cout << str << endl;
    }

    return 0;
}

– Daniel Lim · 3 years, 1 month ago

@Daniel Lim – I don't really think the string library was required but the code pretty much works :D – Charlton Teo · 3 years, 1 month ago

@Charlton Teo – yes, the string library is required or else cin and cout won't work, based on my experience – Daniel Lim · 3 years, 1 month ago

@Daniel Lim – The cin and cout streams are defined in the iostream header and using namespace std is required for their correct interpretation. string header was used for the string data type used which makes the task easier. You can try any of the string ( /character array) input methods like getline, scanf, etc to get the input depending on the type u have defined. – Aditya Raj · 3 years, 1 month ago

@Aditya Raj – yes I can, but I don't really like scanf and I like the syntax of cin more than getline – Daniel Lim · 3 years, 1 month ago

@Daniel Lim – Sure everyone has there own choices but just for fun, once study the scanf features. Its much faster than cin (unless u link cin to scanf). For the part where you dont want to use classes(which I forgot to mention in the previous post :|) Lucas' answer is great as long as you dont want to store the string for future use. 2 simple implementations include creating a char pointer which stores the present string if u need to print it or else if you don't want to store it then just output it during reading only which can be terminated with the delimitter mentioned according to the constraints mentioned or the general delimitter '\0' of strings in c++. – Aditya Raj · 3 years, 1 month ago

@Aditya Raj – yes it is faster indeed, but sometimes it does not work, i dont know – Daniel Lim · 3 years, 1 month ago

@Daniel Lim – why don't you like? – Riaz Ahmad Baboojee · 3 years, 1 month ago

@Daniel Lim – yeaa, I'm sure you can find a way of doing the same thing without using the classes, (therefore using C instead of C++), you could just set up a row counter, read the first parameter to know where to end your loop, and then read the following lines one by one, and when your counter got to a multiple of three you just replicate all the characters you read from input to output and break the line. The coolest way to do it is use the less resources, haha :D – Lucas Franceschi · 3 years, 1 month ago

1	`print("\n".join([input() for _ in range(int(input()))][2::3]))`

One-liner Python, but clearly isn't readable. Basically, range(int(input())) makes a loop of n iterations where n is the number in the input, then [input() for _ in ...] creates a list of the words, iterating n times, then ...[2::3] means taking from the 2nd item (0-based, so 1-based means from the 3rd item) and taking every third item afterwards, then "\n".join(...) means joining all the items, inserting a newline character (\n) between items, and print(...) displays the resulting string.

Let's play code golf! Can you make a program (in any language, but not one specifically designed to solve this particular problem) that solves this in as few characters as possible? The above is 62 characters (61 if you count \n as one character). – Ivan Koswara · 3 years, 1 month ago

Ruby solution: lines = ["7","This", "Is", "Seven","Words", "Or", "More] 0.step(lines.length-1,2) {|index| puts lines [index]} – Viktoriya Forrest · 3 years, 1 month ago

here is the answer:

#include "stdafx.h"
#include <iostream>
#include <string>
#include <string.h>
#include "stdio.h"
#include "string.h"
#include "string"
class Parry
{
    private:
        int n;
        std::string line[100];
    public:
        void getline();
        void code();
};
void Parry::getline()
{
    std::cout<<"Enter the number of lines"<<std::endl;
    std::cin>>n;
    std::cout<<"Enter the lines"<<std::endl;
    for(int i=0;i<n;i++)
    {
        std::cin>>line[i];
    }
}
void Parry::code()
{
    int j=n/3;
    for(int i=1;i<=j;i++)
    {
        std::cout<<line[(3*i)-1]<<std::endl;
    }
}


int _tmain(int argc, _TCHAR* argv[])
{
    Parry p;
    p.getline();
    p.code();
    getchar();
}

– Vaibhav Zambad · 3 years, 1 month ago

JavaScript

function everyThirdLine(input) {
    return input
        .split('\n')
        .reduce(function(memo, n, i) {
            if ((i + 1) % 3 === 0) {
                memo += n + '\n';
            }
            return memo;
        }, '')
        .trim();
}

– J H · 3 years, 1 month ago

You can practice here Hackerrank – Daniel Lim · 3 years, 1 month ago

Here is a pascal solution below: var n,i,j:longint; arr:array [1..1000] of string; begin readln(n); for i:=1 to n do begin readln(arr[i]); end; for j:=1 to n do begin if (j mod 3)=0 then writeln(arr[j]); end; end. – Roihan Munajih · 3 years, 1 month ago

@Roihan Munajih – I tried running this code and they gave me "Error - Program name is missing in main file". Could you be missing something? (Sorry not that familiar with pascal) – Charlton Teo · 3 years, 1 month ago

PHP:

<?PHP
function everyThirdWord($string) {
    $word = explode(" ",$string);
    for($i = 1; $i < str_word_count($string, 0); $i++ ) {
        if($i%3 == 0){
            echo $word[$i-1]."<br>";
        }
    }
}
?>

– Danilo Azevedo · 2 years, 11 months ago

Haskell solution:

-- with IO
main = getLine >> lines <$> getContents >>= \xs -> putStr . unlines $ map (xs !!) [3,6..length xs]
-- without IO (pure)
\xs->map (xs!!)[3,6..length xs]

-- 31 characters. I win!

Why are people learning C++ when Haskell is obviously more awesome? :D – Magnetic Duck · 3 years ago

This is in Microsoft Small Basic :)

count = 0

n = TextWindow.Read()

For i = 1 to n
    word = TextWindow.Read()
    count = count + 1
    If count = 3 Then
        TextWindow.WriteLine(word)
        count = 0
    EndIf
EndFor

– Dana Chiueh · 3 years, 1 month ago

MATLAB

num = inputdlg('Input Number: ','Input');
num = str2num(num);
out =  [];
for x = 1:num
        str = inputdlg('Input Number: ','Input');
        switch _mod(x,3)
              case 0
                    out = [out str '\r\n'];
         end
end
fprintf('%s', out);

– Roi Vinson Abrazaldo · 3 years, 1 month ago

@Roi Vinson Abrazaldo – the 2nd Input Number should have been Input Word. copy/paste mistake – Roi Vinson Abrazaldo · 3 years, 1 month ago

#include <iostream>
#include <string>
using namespace std;

int main() {

    int TC; cin >> TC;
    string str[10]; 
    for(int i = 1; i <= TC; i++)
    {

        cin >> str[i];
    }
      for(int i = 1; i <= TC; i++)
        {
        if(i%3==0)
        cout << str[i] << endl;
          }

    return 0;
}

– Tanmay Sentiya · 3 years, 1 month ago

In JAVA :

import java.util.Scanner;

public class Sample {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner s = new Scanner(System.in);
        int a = s.nextInt();
        String str[] = new String[a];
        for(int i=0;i<a;i++){
            str[i] = s.nextLine();
        }
        for(int i=0;i<a;i++){
            if((i+1)%3==0){
                System.out.println(""+str[i]);
            }
        }
    }

}

– Roopesh H · 3 years, 1 month ago

words=[]
num=int(input("How many words: "))
print("Enter ",num," words..")
while num>0:
        words.append(input())
        num=num-1
num=1
print(words)
for word in words:
    if num == 3:
        print(word)
        num=0
    num=num+1

Not effective Python solution but works. – Shubham Gaikwad · 3 years, 1 month ago

@Shubham Gaikwad – Incorrect, because the problem doesn't ask you to output "How many words:", "Enter x words..", and the list of words. Your output must be precisely as required. – Ivan Koswara · 3 years, 1 month ago

python, using the standard library

1	`output = [x for i, x in enumerate(input_line.split(' ')) if i % 3 is 2]`

– Josh Silverman Staff · 3 years, 1 month ago

Python 3 solution. (Note: this will ONLY work in Python 3 because print is treated as a function and NOT a keyword as Python 2 does):

line_list = [] #List that will store all the lines

input_size = int(input()) #Get all the lines
for i in range(input_size):
    line_list.append(input())

print("") #Adds a new line for readability (optional)

list(map(print, line_list[2::3])) #Print the result

Sample output:

Input:

7
This
Is
Seven
Words
Or
More
Than

Prints:

Seven
More

– L N · 3 years, 1 month ago

Thanks to everyone who tried, anyway here was my solution:

Language: C++

#include<iostream>
using namespace std;
int main(){
string p[1000000];
cin>>x;
for(int i = 1; i <= x; i++){
getline (cin,p[i]);
getline (cin,p[i]);
}
for(int i = 0; i <= x/3; i+=3){
cout<<p[i];
}
}

But i feel that Iamsudip and Daniel's answers were better to please refer to theirs (for mine i didnt need string library i think i use a diff version) – Charlton Teo · 3 years, 1 month ago

Void main() { Int I,n; Char ch[10][10]; Cout<<"enter no of lines"; Cin>>n; For(I=0;I<n;I++) {Cin>>ch[I]; } For(I=0;I<n;I++) { If((i%3==0)&&(I!=0)) Cout<<ch [I]; } Getch (); } – Ashish Kapil · 3 years, 1 month ago

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