Google Code Jam.

google_code_jam/2014/1B-B-lottery.cpp

@@ -0,0 +1,157 @@
+/* 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=""
+*/
+
+/*
+New Lottery Game
+
+The Lottery is changing! The Lottery used to have a machine to generate a random winning number. But due to cheating problems, the Lottery has decided to add another machine. The new winning number will be the result of the bitwise-AND operation between the two random numbers generated by the two machines.
+
+To find the bitwise-AND of X and Y, write them both in binary; then a bit in the result in binary has a 1 if the corresponding bits of X and Y were both 1, and a 0 otherwise. In most programming languages, the bitwise-AND of X and Y is written X&Y.
+
+For example:
+   The old machine generates the number 7 = 0111.
+   The new machine generates the number 11 = 1011.
+   The winning number will be (7 AND 11) = (0111 AND 1011) = 0011 = 3.
+
+With this measure, the Lottery expects to reduce the cases of fraudulent claims, but unfortunately an employee from the Lottery company has leaked the following information: the old machine will always generate a non-negative integer less than A and the new one will always generate a non-negative integer less than B. 
+
+Catalina wants to win this lottery and to give it a try she decided to buy all non-negative integers less than K. 
+
+Given A, B and K, Catalina would like to know in how many different ways the machines can generate a pair of numbers that will make her a winner. 
+
+Could you help her?
+
+Input
+
+The first line of the input gives the number of test cases, T. T lines follow, each line with three numbers A B K.
+
+Output
+
+For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the number of possible pairs that the machines can generate to make Catalina a winner.
+
+Limits
+
+1 ? T ? 100.
+Small dataset
+
+1 ? A ? 1000.
+1 ? B ? 1000.
+1 ? K ? 1000.
+Large dataset
+
+1 ? A ? 109.
+1 ? B ? 109.
+1 ? K ? 109.
+Sample
+
+
+Input 
+ 	
+Output 
+ 
+5
+3 4 2
+4 5 2
+7 8 5
+45 56 35
+103 143 88
+
+Case #1: 10
+Case #2: 16
+Case #3: 52
+Case #4: 2411
+Case #5: 14377
+In the first test case, these are the 10 possible pairs generated by the old and new machine respectively that will make her a winner: <0,0>, <0,1>, <0,2>, <0,3>, <1,0>, <1,1>, <1,2>, <1,3>, <2,0> and <2,1>. Notice that <0,1> is not the same as <1,0>. Also, although the pair <2, 2> could be generated by the machines it wouldn't make Catalina win since (2 AND 2) = 2 and she only bought the numbers 0 and 1.
+*/
+
+#include <assert.h>
+#include <iostream>
+#include <sstream>
+#include <fstream>
+#include <iomanip>
+#include <exception>
+#include <stdexcept>
+#include <map>
+#include <set>
+#include <list>
+#include <vector>
+#include <string>
+#include <memory>
+#include <functional>
+#include <algorithm>
+#include <utility>
+#include <limits>
+#include <math.h>
+
+typedef long long i64;
+typedef std::vector<i64> vec;
+void check( bool b ) { 	if ( !b ) 
+	std::cerr << "error" << std::endl; }
+#define FOR(i,l) for ( i64 i =0, ie = i64(l); i<ie; ++i )
+/*
+	Read n values into v
+*/
+template <class V>
+void readv( V& v, int n )
+{
+	v.reserve(n);
+	for ( int i = 0; i < n; ++i )
+	{
+		typename V::value_type e;
+		std::cin >> e;
+		check( !std::cin.fail() );
+		v.push_back(e);
+	}
+}
+
+
+int solve_puzzle()
+{
+	int a, b, k;
+	std::cin >> a >> b >> k;
+
+	i64 s = 0;
+	for ( int i = 0; i < k; ++i )
+	for ( int i = 0; i < a; i++ )
+		for ( int j = 0; j < b; j++ )
+			if ( (i&j) < k )
+				s++;
+
+	return s;
+}
+
+#undef int
+int main ( void )
+{try{
+	srand((unsigned)time(NULL));
+	int puzzle_count;
+
+	std::cin >>	puzzle_count;
+	std::cin.ignore(std::numeric_limits<std::streamsize>::max(),'\n');
+	for ( int i = 1; i <= puzzle_count; i++ )
+	{
+		std::cout << "Case #" << i << ": ";
+		auto r = solve_puzzle();
+		std::cout << r << 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;
+}}