My solution to the problem A of the round 1A of Google Code Jam 2013

A.py

'''
Problem solution
----------------

The area of a circular ring is:

    pi * (R^2 - r^2)

where R is the radius of the bigger circle and r
the radius of the other one.

Hence, the area of the k-th black ring is (with d
indicating the distance from the innermost circle,
i.e. d = 2k - 1):

    pi * ((r + d)^2 - (r + d - 1)^2) =
  = pi * (2r + 2d - 1)
  = pi * (2r + 2(2k - 1) - 1)
  = pi * (2r + 4k - 3)

If we now sum the areas of N consecutive rings (from k=1):

    pi * (2r + 1)
    pi * (2r + 5)
    pi * (2r + 9)
    ...
    ...

we get:

    N * pi * 2r + pi * (2N^2 - N) =
  = pi * (2Nr + 2N^2 - N) =
  = pi * (2N^2 + (2r - 1)N)

The sequence 1, 5, 9 is http://oeis.org/A016813, and its
sum is http://oeis.org/A000384.

We now equal this to pi * t, which is the maximum area
we can cover with paint:

    pi * (2N^2 + (2r - 1)N) = pi * t
    2N^2 + (2r - 1)N - t = 0

Applying the quadratic formula we finally get:

        |  1 - 2r + sqrt(4r^2 - 4r + 1 + 8t)  |
    N = | ----------------------------------- |
        |_                 4                 _|

where |_x_| is the floor function.
'''

def sqrt(n):
    '''The Newton method applied to sqrt().'''
    x = n
    y = (x + 1) // 2
    while y < x:
        x = y
        y = (x + n // x) // 2
    return x


def rings(r, t):
    '''Returns the maximum number of complete rings one can draw,
    starting with a central circle of radius `r` and `t` milliliters
    of black paint.
    '''
    # We can discard the other solution (the one with the minus) since
    # it would be either a negative value or less than the other one
    # anyway.
    return int((1 - 2 * r + sqrt(4 * r ** 2 - 4 * r + 1 + 8 * t)) // 4)


if __name__ == '__main__':
    import sys
    with open(sys.argv[1], 'w') as out:
        total = int(sys.stdin.readline())
        for i in range(1, total + 1):
            r, t = map(int, sys.stdin.readline().split())
            out.write('Case #{0}: {1}\n'.format(i, rings(r, t)))

Posted one hour after the contest finished, so I hope there won't be problems.

instead of use the formula with sqrt and PI that eliminate itslef:

milliter used = Rn^2 - Rn-1^2
Rn -> circle adius
Rn-1 -> prev circle radius.

R is incremented by 2 (cm)

so is somnething like this:
repeat until t>=0
ml = R^2 * (R-1)^2
.....
r = r+2

very simple.

Isn't that slow on big numbers? I mean, in the large input, r can go up to 10^18 and t to 2 * 10^18. With that method you'll need many iterations to find the final answer. With this method, on the other hand, you just solve a quadratic equation on the integers (so no floating point errors).