tic tac toe

//不觉中 已经全力找工作好久好久了.大概有1年半了.身心疲惫,不要放弃.曙光快来了. 1.tic tac toe //http://www.ntu.edu.sg/home/ehchua/programming/java/JavaGame_TicTacToe.html 类似相关棋牌坐标对战游戏 一通百通 /** * Enumerations for the various states of the game */ public enum GameState { // to save as "G

1. Trees Tree is a recursive structure. 1.1 math nodes https://class.coursera.org/principlescomputing-001/wiki/view? page=trees 1.2 CODE无parent域的树 http://www.codeskulptor.org/#poc_tree.py class Tree: """ Recursive definition for trees plus

题面: C. Team Tic Tac Toe Input ?le: standard input Output ?le: standard output Time limit: 1 second Memory limit: 256 megabytes Farmer John owns 26 cows, which by happenstance all have names starting with different letters of the alphabet, so Farmer J

题目如下: Tic-tac-toe is played by two players A and B on a 3 x 3 grid. Here are the rules of Tic-Tac-Toe: Players take turns placing characters into empty squares (" "). The first player A always places "X" characters, while the second pl

N*N matrix is given with input red or black.You can move horizontally, vertically or diagonally. If 3 consecutive samecolor found, that color will get 1 point. So if 4 red are vertically then pointis 2. Find the winner. def tic_tac_toe(board,n) red,

board = [' ' for i in range(9)] def print_board(): row_1 = '|{}|{}|{}|'.format(board[0], board[1], board[2]) row_2 = '|{}|{}|{}|'.format(board[3], board[4], board[5]) row_3 = '|{}|{}|{}|'.format(board[6], board[7], board[8]) print('\n') print(row_1)

题目:给你一个井字棋的状态,判断是否合法. 分析:枚举.直接枚举多有情况判断即可. 合法状态有三种情况:(X先下子) 1.X赢,则O不能赢,且X比O多一子: 2.O赢,则X不能赢,且O和X子一样多: 3.没人赢,此时O的子和可能和X一样多,也可能少一个. 说明:简单题(⊙_⊙). #include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include &l

17.2 Design an algorithm to figure out if someone has won a game oftic-tac-toe. 这道题让我们判断玩家是否能赢井字棋游戏,有下面几点需要考虑: 1. 判断是否能赢hasWon函数是调用一次还是多次,如果是多次,我们可能为了优化而需要加入一些预处理. 2. 井字棋游戏通常是3x3的大小,我们是否想要实现NxN的大小? 3. 我们需要在代码紧凑,执行速度和代码清晰之间做出选择. #include <iostream> #

我的代码: #include <bits/stdc++.h> using namespace std; int main() { char a[3][3]; int i,j=0; for(i=0;i<3;i++){ for(j=0;j<3;j++){ cin>>a[i][j]; } } int count1,count2; int len1,len2; len1=len2=0; char b[50],c[2][50]; count1=count2=0; for( i=0