This is the O(1) solution based on the pattern.

for _ in xrange(input()):
     print ["First","Second"][input()%7 in [0,1]]

Guys try it yourself for like 20 cases, on a paper! You will find out :P

The pattern emerges pretty quickly if you start from 1 stone and look at if the first player will win or lose with each stone you add.

SPOILER

You'll find that the first player will lose when the starting number of stones s is: 1,7,8,14,15,21,22... (i.e. when s % 7 < 2). With this knowledge in hand the code is trivial:

t = int(raw_input())
for _ in xrange(t):
    s = int(raw_input())
    if s % 7 < 2:
        print 'Second'
    else:
        print 'First'

My solution

int player2(int n){

if(n<2)
    return 1;

return (player1(n-5)&&player1(n-3)&&player1(n-2));

}

int player1(int n){

if(n<2)
    return 0;

return (player2(n-5)||player2(n-3)||player2(n-2));

}

int main() {

int n,t,i,j;
scanf("%d",&t);
for(int a0=0;a0<t;++a0){

    scanf("%d",&n);

    i = player1(n);

    if(i==1)
        printf("First\n");
    else
        printf("Second\n");

}

return 0;

}

This game is composed of the repeating group of LWWWWWL Therefore the solution in Java is like this --> System.out.println(scanner.nextInt()%7<2?"Second":"First");

Source: https://www.cs.umd.edu/~gasarch/COURSES/250/S15/nimnotes.pdf

Is there any actual theory behind this game?

Any other way to deduce the logic instead of just putting the numbers in and seeing the pattern?

Thanks, Sidpat

The key to solve this problm is enumerate it out on a paper.

At first I thought wow when the n is big there were a lot of possibilities so how could it possible to have a fated loss or win. I was wrong.

C++ Solution and Explaination:

At first I had tried to modulos N by 10 (n = n%10) and set the number of if statements for each possibility from 1 to 10 to determine the winner, I was completely wrong. So I purchased the Test Cases for TC #0 and noticed that the pattern repeated itself after the 7th entry, it would go "Second First First First First First Second" and after that the pattern was the same. So I realized the best answer was to check if n was equal or greater than 7 and if so modulus n by 7 (n = n % 7). If the remainder was 0, it meant n was equal to the 7th case of the pattern (7 or 14 for example) and Player 2 Wins or if the reminder was 1, it meant n was equal to the 1st case of the pattern (8 or 15 for example) and Player 2 wins, any other remainder meant Player 1 wins.

int main() {
    int t;
    cin >> t;
    
    for(int i = 0; i < t; i++){
        int n;
        cin >> n;
        
        if(n >= 7)
            n %= 7;
        
        if(n == 0 || n == 1)
            cout << "Second" <<  endl;
        else
            cout << "First" << endl;
        
    }
    return 0;
}

Tyipical case of dynamic programing.

This problem is biased to player one. if Any of the move chain exists which lead to player one to win player one to win.

Surprised to see this marked as easy when a similar problem was marked as hard in another contest.