该楼层疑似违规已被系统折叠 隐藏此楼查看此楼

二楼代码贴上

#include

#define Esc 27

#define Enter 13

#define Wall 1

#define Road 0

#define Fruit 2

#define Yaoshi 3

#define Rukou 4

#define Chukou 5

#define Door 6

#define Monster 7

#define Life 8

#define UP 0x48

#define DOWN 0x50

#define LEFT 0x4b

#define RIGHT 0x4d

map[20][60]={{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},

{1,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,7,7,1,1,1,1,1,1,1,1,1,1,1},

{1,0,0,0,0,0,0,0,0,0,0,0,0,7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,0,8,0,0,0,7,0,0,0,8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,1,1,1,1,1,6,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,3,7,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,2,2,0,0,0,7,0,0,0,0,2,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,1,1,1,1,1,6,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,1},

{1,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,8,3,1,0,0,0,0,0,1},

{1,0,0,0,0,1,0,0,1,0,0,0,0,1,1,1,1,6,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,6,7,0,0,0,0,0,0,0,0,0,0,0,6,7,0,0,2,8,1,0,0,0,0,0,1},

{1,2,2,0,0,1,0,0,1,0,0,2,2,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,8,3,1,0,0,0,0,0,1},

{1,8,2,0,0,7,0,0,7,0,0,2,8,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,1,1,1,1,1,1,1,1,6,1,1,1,1,1,1,0,0,0,0,0,1},

{1,2,2,0,0,1,0,0,1,0,0,2,2,1,1,1,1,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,0,0,0,0,1,0,0,1,0,0,0,0,1,2,2,0,0,0,2,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,7,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},

{1,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,7,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1},

{1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,6,0,0,0,7,2,2,8,3,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,6,0,0,0,0,0,0,0,5,0,1},

{1,1,2,3,8,0,7,7,0,8,3,2,1,1,8,3,7,0,7,3,8,1,0,0,0,7,2,2,8,3,1,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1},

{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},};

maplook()

{

int i,j,k;

for(i=0;i<20;i++)

{

for(j=0;j<60;j++)

{

switch(map[i][j])

{

case Wall:textcolor(10),cprintf("%c",254);break;

case Road:printf(" ");break;

case Fruit:textcolor(14),cprintf("%c",12);break;

case Yaoshi:textcolor(9),cprintf("%c",15);break;

case Rukou:textcolor(14),cprintf("%c",175);break;

case Chukou:textcolor(14),cprintf("%c",174);break;

case Door:textcolor(9),cprintf("%c",176);break;

case Monster:textcolor(12),cprintf("%c",234);break;

case Life:textcolor(13),cprintf("%c",3);break;

}

}

printf("\n");

}

}

game_move()

{

int i,j,x=2,y=2,yaoshi=0,life=50,grade=0;

char c,c1;

do

{

c=getch();

switch(c)

{

case UP:y--;

if(map[y-1][x-1]==1||((map[y-1][x-1]==6)&&yaoshi==0))

{

y++;

}

break;

case DOWN:y++;

if(map[y-1][x-1]==1||((map[y-1][x-1]==6)&&yaoshi==0))

{

y--;

}

break;

case LEFT:x--;

if(map[y-1][x-1]==1||((map[y-1][x-1]==6)&&yaoshi==0))

{

x++;

}

break;

case RIGHT:x++;

if(map[y-1][x-1]==1||((map[y-1][x-1]==6)&&yaoshi==0))

{

x--;

}

break;

case Esc:exit(1);

}

if(map[y-1][x-1]==Fruit)

{

map[y-1][x-1]=0;

grade+=10;

}

if(map[y-1][x-1]==Yaoshi)

{

map[y-1][x-1]=0;

yaoshi++;

}

if(map[y-1][x-1]==Door)

{

map[y-1][x-1]=0;

yaoshi--;

}

if(map[y-1][x-1]==Monster)

{

map[y-1][x-1]=0;

life-=10;

grade+=20;

if(life<=0)

{

gotoxy(68,17),textcolor(12);cprintf("Game Over");

getch();

exit(1);

}

}

if(map[y-1][x-1]==Life)

{

map[y-1][x-1]=0;

life+=10;

}

clrscr();

maplook();

textcolor(14);

gotoxy(68,11),cprintf("keys:%c %d",15,yaoshi);

gotoxy(68,13),cprintf("life:%c %d",3,life);

gotoxy(68,15),cprintf("grade%c %d",14,grade);

gotoxy(x,y),textcolor(15),cprintf("%c\b",1);

}while(1);

}

main()

{

maplook();

game_move();

getch();

}

C语言魔塔视频百度云,原创 C语言版魔塔相关推荐

  1. 北华大学c语言题库百度云,北华大学C语言题库精简打印版(全).doc

    北华大学C语言题库精简打印版(全).doc 北华大学C语言题库精简打印版(全)一.判断题 - 正确篇1.字符常量的长度肯定为1.Y2.在调用函数时,实参把值传送给对应位置上的形参,形参的值不能传给实参 ...

  2. c语言深度剖析百度云,《C语言深度剖析》笔记

    <C语言深度剖析>笔记 1.在c语言中,凡不加返回值类型限定的函数,就会被编译器作为返 回整形处理. 2.register 变量必须是一个单个的值,并且其长度应小于或等于整型 的长度.而且 ...

  3. python百度云资源-Python开发视频百度云分享

    原标题:Python开发视频百度云分享 Python有很好的3D渲染库和游戏开发框架,有很多使用Python开发的游戏,如迪斯尼卡通城.黑暗之刃.常用PyGame.Pykyra等和一个PyWeek的比 ...

  4. python视频免费百度云-Python开发视频百度云分享

    原标题:Python开发视频百度云分享 Python有很好的3D渲染库和游戏开发框架,有很多使用Python开发的游戏,如迪斯尼卡通城.黑暗之刃.常用PyGame.Pykyra等和一个PyWeek的比 ...

  5. c语言程序设计第三版百度云,c语言编程练习题百度云.doc

    c语言编程练习题百度云 精品文档2016全新精品资料-全新公文范文-全程指导写作 –独家原创 PAGE1 / NUMPAGES1 c语言编程练习题百度云 共花了50先令,每个男人各花3先令,每个女人各 ...

  6. 千锋中级Python视频百度云

    千锋中级Python视频百度云免费赠送给大家,希望大家在Python学习路上一帆风顺! 所属网站分类: 资源下载 > python视频教程 作者:外星人入侵 原文链接: http://www.p ...

  7. mysql动力节点百度云_动力节点MySQL数据库视频 百度云 网盘 下载

    资源名称:动力节点MySQL数据库视频 百度云 网盘 下载 # o4 E. q% ]2 ?百度网盘下载链接:[/hide]- t4 L+ S# b2 T( }! d) n& d 密码:dff7 ...

  8. 西邮c语言期末试卷百度云,GitHub - dcfun/XiyouLibNodeExpress: 西邮图书馆Web API-Node.js...

    XiyouLibNodeExpress 该API现已由胖萌维护,如有问题,请联系:910739015@qq.com 基于Node.js Express框架的西邮图书馆REST API 请根据下面的介绍 ...

  9. 百度云不限速c语言,如何解决百度云下载大文件限速问题

    最痛苦的事情莫过于下载一个N个G的文件,而速度只有80k/s. 有一个东西叫百度云会员 我想,大部分人都很痛恨下载限速吧?尤其是当360云关门大吉后,百度云就可以更加嚣张的为所欲为了.不开百度云会员, ...

最新文章

  1. visual studio怎么重启?(visual studio restart插件)
  2. 贝叶斯优化神经网络参数_贝叶斯超参数优化:神经网络,TensorFlow,相预测示例
  3. 住150平米以上的房子是怎样一种体验?
  4. 中科院发布了目标追踪数据集,1万多条视频,150万个边界框 | 快来下载
  5. SpringBoot-文件在线预览解决方案-基于OpenOffice及jacob
  6. stm32中断优先级_浅谈STM32串口USART1的使用
  7. 计算机专业评定职称论文,《计算机职称论文.doc
  8. one loop per thread
  9. 博途编程语言切换_从一种编程语言切换到另一种:灵活的好处
  10. 学习日记day29 平面设计 色彩
  11. 从5点来分析搜索引擎算法
  12. linux下view如何修改字符串,Linux下view命令的使用
  13. 使用ScanShadowsFilter过滤激光雷达拖尾
  14. win7 系统装SQLServer2000 成功。
  15. 批处理备份及删除,forfiles命令详解
  16. 关于参加大学生数学竞赛的一点感悟与体会
  17. Android安卓系统提示应用程序未安装的解决方法
  18. IDEA打可执行jar包详细教程(包含依赖的所有jar包)
  19. 配置管理之二变更管理
  20. Linux-2.6.32.2内核在mini2440上的移植(四)---根文件系统制作(1)

热门文章

  1. 六轴机械臂控制原理图_六轴工业机器人工作原理解析
  2. 【新书推荐】【2018】战术长时监视雷达及其应用
  3. 485与计算机连接,RS422与RS485的连接方法
  4. 火车头采集器 v9免费版使用
  5. Apple watch实现Personal Scrum
  6. 公有云上基于微服务架构SAAS产品研发实践「活动通知」
  7. C++程序设计基础案例教程pdf
  8. Linux系统ORACLE 19C OEM监控管理
  9. vc获取计算机用户名,vc获取计算机名和ip地址的方法
  10. 实时音视频直播新玩法中的混音技术