当前位置: 首页 > ds >正文

2048小游戏

熟悉的2048小游戏,环境:Windows10及以上,dev.c++5.11及以上或控制台(可能有警告,不影响运行)

付代码:

#include <iostream>
#include <cstdlib>
#include <ctime>using namespace std;const int SIZE = 4;
int board[SIZE][SIZE];// 初始化游戏棋盘,随机两个位置生成 2 或 4
void initialize() {for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {board[i][j] = 0;}}srand(time(0));int x1 = rand() % SIZE;int y1 = rand() % SIZE;board[x1][y1] = (rand() % 2 == 0) ? 2 : 4;int x2, y2;do {x2 = rand() % SIZE;y2 = rand() % SIZE;} while (x2 == x1 && y2 == y1);board[x2][y2] = (rand() % 2 == 0) ? 2 : 4;
}// 打印游戏棋盘
void printBoard() {system("cls");for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {if (board[i][j] == 0) {cout << ".\t";} else {cout << board[i][j] << "\t";}}cout << endl;}
}// 在空白位置随机生成一个 2 或 4
void generateRandomTile() {int emptyTiles = 0;for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {if (board[i][j] == 0) {emptyTiles++;}}}if (emptyTiles > 0) {int randomTile = rand() % emptyTiles;int count = 0;for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {if (board[i][j] == 0) {if (count == randomTile) {board[i][j] = (rand() % 2 == 0) ? 2 : 4;return;}count++;}}}}
}// 判断是否还有空白位置
bool hasEmptyTile() {for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {if (board[i][j] == 0) {return true;}}}return false;
}// 判断是否还有可合并的相邻方块
bool canMerge() {for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {if (i > 0 && board[i][j] == board[i - 1][j]) {return true;}if (i < SIZE - 1 && board[i][j] == board[i + 1][j]) {return true;}if (j > 0 && board[i][j] == board[i][j - 1]) {return true;}if (j < SIZE - 1 && board[i][j] == board[i][j + 1]) {return true;}}}return false;
}// 判断游戏是否结束
bool isGameOver() {return!hasEmptyTile() &&!canMerge();
}// 向上移动
void up() {bool moved = false;for (int j = 0; j < SIZE; ++j) {for (int i = 1; i < SIZE; ++i) {if (board[i][j] != 0) {int k = i;while (k > 0 && board[k - 1][j] == 0) {board[k - 1][j] = board[k][j];board[k][j] = 0;k--;moved = true;}if (k > 0 && board[k - 1][j] == board[k][j]) {board[k - 1][j] *= 2;board[k][j] = 0;moved = true;}}}}if (moved) {generateRandomTile();}
}// 向下移动
void down() {bool moved = false;for (int j = 0; j < SIZE; ++j) {for (int i = SIZE - 2; i >= 0; --i) {if (board[i][j] != 0) {int k = i;while (k < SIZE - 1 && board[k + 1][j] == 0) {board[k + 1][j] = board[k][j];board[k][j] = 0;k++;moved = true;}if (k < SIZE - 1 && board[k + 1][j] == board[k][j]) {board[k + 1][j] *= 2;board[k][j] = 0;moved = true;}}}}if (moved) {generateRandomTile();}
}// 向左移动
void left() {bool moved = false;for (int i = 0; i < SIZE; ++i) {for (int j = 1; j < SIZE; ++j) {if (board[i][j] != 0) {int k = j;while (k > 0 && board[i][k - 1] == 0) {board[i][k - 1] = board[i][k];board[i][k] = 0;k--;moved = true;}if (k > 0 && board[i][k - 1] == board[i][k]) {board[i][k - 1] *= 2;board[i][k] = 0;moved = true;}}}}if (moved) {generateRandomTile();}
}// 向右移动
void right() {bool moved = false;for (int i = 0; i < SIZE; ++i) {for (int j = SIZE - 2; j >= 0; --j) {if (board[i][j] != 0) {int k = j;while (k < SIZE - 1 && board[i][k + 1] == 0) {board[i][k + 1] = board[i][k];board[i][k] = 0;k++;moved = true;}if (k < SIZE - 1 && board[i][k + 1] == board[i][k]) {board[i][k + 1] *= 2;board[i][k] = 0;moved = true;}}}}if (moved) {generateRandomTile();}
}int main() {initialize();while (true) {printBoard();if (isGameOver()) {cout << "Game Over!" << endl;break;}char input;cin >> input;switch (input) {case 'w':up();break;case 's':down();break;case 'a':left();break;case 'd':right();break;default:cout << "请输入'w', 's', 'a',d'" << endl;}}return 0;
}

http://www.xdnf.cn/news/17235.html

相关文章:

  • 【java】大数据insert的几种技术方案和优缺点
  • (ZipList入门笔记一)ZipList的节点介绍
  • Windows 电脑远程访问,ZeroTier 实现内网穿透完整指南(含原理讲解)
  • Spring Boot 整合 Web 开发全攻略
  • 深度拆解Dify:开源LLM开发平台的架构密码与技术突围
  • 消息队列疑难问题(RocketMQ)
  • 09-堆
  • GaussDB 常见问题-集中式
  • 8.5 CSS3多列布局
  • lumerical——Y分支功分器
  • Redis Stream:高性能消息队列核心原理揭秘
  • PDF转图片工具技术文档(命令行版本)
  • CRT调试堆检测:从原理到实战的资源泄漏排查指南
  • 北京JAVA基础面试30天打卡02
  • RocketMq如何保证消息的顺序性
  • 面向对象的七大设计原则
  • Kotlin属性委托
  • 探秘MOBILITY China 2026,新能源汽车与智慧出行的未来盛宴
  • React18 严格模式下的双重渲染之谜
  • 嵌入式硬件中运放的基本控制原理
  • 2025金九银十Java后端面试攻略
  • 天津大学2024-2025 预推免 机试题目(第二批)
  • 400V降24V,200mA,应用领域:从生活到工业的 “全能电源管家”
  • C++面向对象编程基础:从类定义到封装机制详解
  • 深度学习-卷积神经网络CNN-填充与步幅
  • 最新基于Python科研数据可视化实践技术
  • 【人工智能99问】什么是Post-Training,包含哪些内容?(19/99)
  • Next Terminal 实战:内网无密码安全登录
  • MCP进阶:工业协议与AI智能体的融合革命
  • Redis之Hash和List类型常用命令