Move Google Code Jam under puzzles.

puzzles/google_code_jam/2013/1A-A-bullseye.cpp

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+/* Check cf5-opt.vim defs.
+VIM: let g:lcppflags="-std=c++11 -O2 -pthread"
+VIM: let g:wcppflags="/O2 /EHsc /DWIN32"
+VIM: let g:cppflags=g:Iboost.g:Itbb
+VIM: let g:ldflags=g:Lboost.g:Ltbb.g:tbbmalloc.g:tbbmproxy
+VIM: let g:ldlibpath=g:Bboost.g:Btbb
+VIM: let g:argv=" < tic_tac_toy_tomek.sample_input"
+*/
+/*!
+Problem A. Bullseye
+This contest is open for practice. You can try every problem as many times as you like, though we won't keep track of which problems you solve. Read the Quick-Start Guide to get started.
+Small input
+11 points	
+Solve A-small
+Large input
+13 points	
+Solve A-large
+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 black-and-white 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 ? cm2. One millilitre of paint is required to cover area ? cm2. 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 ? 1018.
+1 ? t ? 2 ? 1018.
+
+Sample
+
+
+Input 
+ 	
+5
+1 9
+1 10
+3 40
+1 1000000000000000000
+10000000000000000 1000000000000000000
+
+
+Output 
+ 
+Case #1: 1
+Case #2: 2
+Case #3: 3
+Case #4: 707106780
+Case #5: 49
+*/
+
+#include <iostream>
+#include <exception>
+#include <vector>
+#include <string>
+#include <stdexcept>
+#include <math.h>
+#include <assert.h>
+
+
+long long solve_puzzle()
+{
+	long long r;
+	long long t;
+	std::cin >> r;
+	std::cin >> t;
+	
+	long long a = 2;
+	long long b = (2*r-1);
+	long long c = -t;
+//	long double det = double(b)*b-4*a*c;
+//	long double root = (sqrt(det)-b)/(2*a);
+//	long long n = (long long)(floor(root));
+
+	long long n = 1;
+	while ( a*n*n+b*n <= t )
+		n *=2;
+
+	long long l = 1;
+	long long h = n;
+	while ( h != l+1 )
+	{
+		long long m = (h+l)/2;
+		if ( a*m*m+b*m <= t)
+			l = m;
+		else
+			h = m;
+	}
+
+	if ( a*l*l+b*l > t || a*h*h+b*h <= t )
+		std::cerr << "bug" << std::endl;
+
+	return l;
+}
+
+int main ( void )
+{try{
+
+	int puzzle_count;
+
+	std::cin >>	puzzle_count;
+	std::cin.ignore(std::numeric_limits<std::streamsize>::max(),'\n');
+	for ( int i = 0; i < puzzle_count; i++ )
+	{
+		auto s = solve_puzzle();
+		std::cout << "Case #" << (i+1) << ": " <<s << std::endl;
+	}
+
+	return 0;
+}
+catch ( const std::exception& e )
+{
+	std::cerr << std::endl
+			<< "std::exception(\"" << e.what() << "\")." << std::endl;
+	return 2;
+}
+catch ( ... )
+{
+	std::cerr  << std::endl
+			<< "unknown exception." << std::endl;
+	return 1;
+}}
+