// Greetings, fellow coders!! I welcome you to a new program of mine. It's supposed // to check a solution for a problem. (as you may know, that's something I really like to do) // I will tell you the problem, but the solution is yours to figure out, even though the comments // are almost giving it away... ///////////////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////////////////////////////////// // We start with the number 2 written on the board. Two folks are playing a game that in each // person's turn, they must add to the number on the board, one of its divisors besides itself. // and write it on the board. The winner is the one who writes a number greater than 1990 on the board. // Now, what's the "winning strategy" of the game? #include #include #include #include #include using namespace std; // function to factorize an integer into all its positive divisors besides itself vector factorize(int n) { vector factors; for (int iter = 1; iter <= n / 2; iter++) { if (n % iter == 0) factors.push_back(iter); } return factors; } int main() { // to count how many times each player has won the game int firstWon = 0, secondWon = 0; // loop that plays the game for 1,000 times for (int j = 0; j < 1e3; j++) { // num is the number written on the board, and i shows whose turn it is int num = 2, i = 0; // a loop that stops the game when a number greater than 1990 is written on the board while (num < 1990) { // randomize seed ( there is a "+ rand()" here, because the loop happens quite a lot of times in one second, so if it were not there, we'd get the same random numbers for each second the program worked which means a lot of loop executions giving out the same results, kay? ) srand(time(NULL) + rand()); // factorize the number written on the board vector factors = factorize(num); // if the sum of the number on the board and its greatest divisor is greater than 1990, win (would happen for both player #1 and player #2, the catch is that player #2 never gets the chance!) if (num + factors.back() > 1990) num += factors.back(); // if it's player #2's choice add a random divisor (I mean, what other choice do they have?) if (i % 2 == 1) { num += factors[rand() % factors.size()]; i++; } // if not, add 1, since 1 is an odd divisor of all integers else { num += 1; i++; } } // if the game ends in odd turns, add to first player's wins if (i % 2 == 1) { firstWon++; } // if it ends in even turns, add to second player's wins else { secondWon++; } } cout << firstWon << " : " << secondWon << " ---> first player to second player " << endl; // display the end result // pause for input system("pause"); return 0; } // if you've come this far, then I suggest you change 1990 (on lines 49 and 58) to another number and see what happens (what if you tweak the turns as well?)

Problem Solution Checker # 2