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;
}

Here is a python2 solution below:

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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 

C++

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#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;
}

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).

Ruby solution: lines = ["7","This", "Is", "Seven","Words", "Or", "More] 0.step(lines.length-1,2) {|index| puts lines [index]}

here is the answer:

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#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();
} 

JavaScript

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

You can practice here Hackerrank

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.

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>";
        }
    }
}
?>

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

This is in Microsoft Small Basic :)

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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

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);

In JAVA :

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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]);
            }
        }
    }

} 

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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.

python, using the standard library

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

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

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

Language: C++

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#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)

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 (); }