Most of us are familiar with google codejam, those who don’t visit https://code.google.com/codejam for more information. This is the first problem from Online Round 1A 2013, top 1000 from this round will be eligible for next online round.
Problem
Maria has been hired by the Ghastly Chemicals Junkies (GCJ) company to help them manufacture bullseyes. A bullseye consists of a number of concentric rings (rings that are centered at the same point), and it usually represents an archery target. GCJ is interested in manufacturing blackandwhite bullseyes.
Maria starts with t millilitres of black paint, which she will use to draw rings of thickness 1cm (one centimetre). A ring of thickness 1cm is the space between two concentric circles whose radii differ by 1cm.
Maria draws the first black ring around a white circle of radius r cm. Then she repeats the following process for as long as she has enough paint to do so:
 Maria imagines a white ring of thickness 1cm around the last black ring.
 Then she draws a new black ring of thickness 1cm around that white ring.
Note that each “white ring” is simply the space between two black rings.
The area of a disk with radius 1cm is π cm^{2}. One millilitre of paint is required to cover area π cm^{2}. What is the maximum number of black rings that Maria can draw? Please note that:
 Maria only draws complete rings. If the remaining paint is not enough to draw a complete black ring, she stops painting immediately.
 There will always be enough paint to draw at least one black ring.
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of a line containing two space separated integers: r and t.
Output
For each test case, output one line containing “Case #x: y“, where x is the case number (starting from 1) and y is the maximum number of black rings that Maria can draw.
Limits
Small dataset
1 ≤ T ≤ 1000.
1 ≤ r, t ≤ 1000.
Large dataset
1 ≤ T ≤ 6000.
1 ≤ r ≤ 10^{18}.
1 ≤ t ≤ 2 × 10^{18}.
Sample
Input  
5 

Output  

Solution with Java Code and Explanation
First we will calculate the area of the first black circle = π(r+1)^^{2} – πr^^{2} = π(2r+1);
Similarly area of the second black circle = π(r+1+2)^^{2} – π(r+1+1)^^{2} = π(2r+5);
Similarly area of the third black circle = π(r+1+4)^^{2} – π(r+1+3)^^{2} = π(2r+9);
This will form an AP with T_{n} = π((2r+1)+ (n1)*4);
Sum of this AP S_{n}= n(a1 + an)/2 = nπ(2r+1 + 2r +1 + 4n 4)/2 = nπ(4r + 2 + 4n 4)/2 = π(2rn + 2n^^{2} n)
We need to solve this sum for t => 2n^^{2} + n(2r1) t = 0
We can use quadratic solution for any ax^^{2} + bx +c = 0 = (b +/ sqrt(b^^{2} – 4ac))/2, we can reject – options as it will become negative.
here a = 2, b = 2r 1 and c = t
NOTE: for writing the java code we have to use biginteger as the rang is too high and main problem you might face is to calculate square root of biginteger in java, this sample code below includes how you can calculate square root of biginteger in java.
Here is the Java code for Bullseye problem in Google codejam Online Round 1A 2013
package googlecodejam2013; import java.io.BufferedReader; import java.io.File; import java.io.FileReader; import java.io.FileWriter; import java.io.Reader; import java.math.BigInteger; public class Bullseye { private static final BigInteger TWO = BigInteger.valueOf(2); private static boolean isSqrt(BigInteger n, BigInteger root) { final BigInteger lowerBound = root.pow(2); final BigInteger upperBound = root.add(BigInteger.ONE).pow(2); return lowerBound.compareTo(n) <= 0 && n.compareTo(upperBound) < 0; } public static BigInteger sqrt(BigInteger n) { if (n.signum() >= 0) { final int bitLength = n.bitLength(); BigInteger root = BigInteger.ONE.shiftLeft(bitLength / 2); while (!isSqrt(n, root)) { root = root.add(n.divide(root)).divide(TWO); } return root; } else { throw new ArithmeticException("square root of negative number"); } } public static String getCircleCount(BigInteger r, BigInteger t){ BigInteger b = r.multiply(new BigInteger("2")).subtract(new BigInteger("1")); BigInteger a = new BigInteger("2"); BigInteger ans = b.multiply(b).add(t.multiply(new BigInteger("8"))); ans = sqrt(ans).subtract(b).divide(new BigInteger("4")); return ans.toString(); } public static void main(String[] argv) { try { long startTime = System.currentTimeMillis(); Reader reader = new FileReader("sample.in"); BufferedReader bufReader = new BufferedReader(reader); String x = bufReader.readLine(); int numOfTestCases = Integer.parseInt(x); int count = 0; File file = new File("sample.out"); FileWriter writer = new FileWriter(file); for(count =1; count<= numOfTestCases; count++) { String line = bufReader.readLine(); BigInteger r = new BigInteger(line.substring(0, line.indexOf(" "))); BigInteger t = new BigInteger(line.substring( line.indexOf(" ")+1)); String output = getCircleCount(r, t); writer.write("Case #" + count + ": " + output+"\n"); System.out.println("Case #" + count + ": " + output); } writer.close(); System.out.println("Total time = " + (System.currentTimeMillis()  startTime)); } catch (Exception e) { e.printStackTrace(); } } }
Altamir Gomes Bispo Junior says
Another suggestion for avoiding overflows is to employ the bisection method with doable ranges.
Leandro says
That’s ok. I got it.
Leandro says
Hi,
Ins’t the formula for circle area equals PI * (r)^2? Why is it necessary to subtract, like π(r+1)^2 – πr^2
Thank you.
Kthik says
Because the area of the black paint isnt a full circle. It’s the difference between two circles.