SPOJ #752. Power it!

By property of mod operations , we can simply use Divide and Conquer + Recursion to solve it. Reference: https://www.khanacademy.org/math/applied-math/cryptography/modarithmetic/a/modular-exponentiation

My Ruby version is:

DIV = 20

def ferma(x, y, n)
    c = 1
    for i in 0..y-1
        c = (c * x) % n         
    end
    #p "[#{x}^#{y} mod #{n} = #{c}]"
    return c
end

def div_conq_ferma(x, y, n)    
    #p "#{x}^#{y} mod #{n} = (#{x}^#{DIV})^#{y/DIV} * #{x}^#{y%DIV}, mod #{n}"

    mod1 = ferma(x, y % DIV, n)

    if (y > DIV)
        sub_mod0 = ferma(x, DIV, n)
        pwr_sub_mod0 = y / DIV
        mod0 = div_conq_ferma(sub_mod0, pwr_sub_mod0, n)
    else
        mod0 = 1
    end
    
    return ferma(mod1 * mod0, 1, n)
end
#
runcnt = gets.to_i
for i in 0..runcnt-1
    str = gets.split
    x = str[0].to_i
    y = str[1].to_i
    n = str[2].to_i    
    p div_conq_ferma(x, y, n)
end

But seems Ruby is not fast enough to pass the second case. All Ruby submissions failed the 2nd case with TLE.