2021年推箱子C课程设计报告

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推箱子C语言程序报告

推箱子C语言程序报告推箱子是一种极富智慧和策略性的游戏，是著名的益智游戏之一、它不仅可以锻炼人们的逻辑思维能力和创造力，而且还能提高人们的解决问题的能力和耐心。

本篇报告将介绍一个基于C语言开发的推箱子游戏，包括游戏规则、程序设计思路和实现效果等方面的内容。

一、游戏规则推箱子游戏的规则十分简单，玩家需要操作一个小人将该场景中的木箱全部推到指定位置。

在游戏开始时，场景中会放置一定数量的木箱和一个小人。

玩家可以通过键盘输入上下左右四个方向键来移动小人，小人可以推动场景中的木箱，但不能直接拉动或推动多个箱子。

当所有的木箱都被推到指定位置时，游戏胜利，反之则失败。

二、程序设计思路推箱子游戏的实现涉及到场景的渲染、小人和木箱的移动、游戏的逻辑判断等多个方面。

下面将分别介绍这些方面的程序设计思路。

1.场景渲染2.小人和木箱的移动玩家操作小人的移动，可以通过监听键盘输入的方向键来实现。

根据输入的方向，判断小人与目标位置的关系，如果目标位置是空地或者指定位置，则小人可以移动到目标位置，否则不能移动。

当小人移动后，如果目标位置是木箱，则需要判断木箱与目标位置的关系，如果目标位置是空地或者指定位置，则木箱可以被推动到目标位置，否则不能推动。

3.游戏逻辑判断在每一次小人移动后，需要判断游戏是否胜利或失败。

胜利的条件是所有的木箱都被推到指定位置，判断的方法是在遍历整个场景时，检查是否存在未推到指定位置的木箱。

失败的条件是小人无法再移动，判断的方法是在判断小人是否能够移动时，如果没有可移动的方向，则游戏失败。

三、实现效果通过以上的程序设计思路，可以开发出一个基于C语言的推箱子游戏。

该游戏可以提供丰富的关卡和难度选择，让玩家能够不断挑战和提高自己的智力。

游戏的界面简洁明了，操作方便快捷，给玩家带来舒适的游戏体验。

总结：推箱子是一种极富智慧的益智游戏，本篇报告介绍了一个基于C语言开发的推箱子游戏。

通过对游戏规则、程序设计思路和实现效果的阐述，可以看出该游戏具有多样化的场景和难度选择，能够锻炼玩家的智力和思维能力。

c课程设计推箱子

c 课程设计推箱子一、教学目标本课程的教学目标是使学生掌握推箱子游戏的算法和编程技巧，培养学生的逻辑思维能力和问题解决能力。

具体目标如下：知识目标：使学生了解推箱子游戏的基本概念和算法，理解游戏编程的基本原理。

技能目标：培养学生能够独立完成推箱子游戏的编程和调试，提高学生的编程实践能力。

情感态度价值观目标：培养学生对计算机科学的兴趣和热情，培养学生的创新精神和团队合作意识。

二、教学内容本课程的教学内容主要包括推箱子游戏的基本概念、算法和编程技巧。

具体安排如下：第一章：推箱子游戏概述，介绍推箱子游戏的基本概念和特点。

第二章：推箱子游戏的算法，讲解推箱子游戏的解决方法和算法实现。

第三章：推箱子游戏的编程，介绍推箱子游戏的编程语言和编程技巧。

第四章：推箱子游戏的调试和优化，讲解如何对推箱子游戏进行调试和优化。

三、教学方法本课程的教学方法主要包括讲授法、案例分析法和实验法。

具体使用如下：讲授法：用于讲解推箱子游戏的基本概念、算法和编程技巧。

案例分析法：通过分析具体的推箱子游戏案例，使学生更好地理解和掌握相关知识。

实验法：让学生通过实际编程和调试推箱子游戏，提高学生的实践能力。

四、教学资源本课程的教学资源主要包括教材、参考书、多媒体资料和实验设备。

具体使用如下：教材和参考书：为学生提供推箱子游戏的基本概念、算法和编程技巧的学习资料。

多媒体资料：通过视频、动画等形式，为学生提供更为直观的学习资源。

实验设备：为学生提供实际编程和调试推箱子游戏的硬件支持。

五、教学评估本课程的评估方式包括平时表现、作业和考试三个部分，以保证评估的客观性和公正性。

平时表现评估：通过观察学生在课堂上的参与程度、提问回答和小组讨论表现，评估学生的学习态度和理解能力。

作业评估：通过布置相关的编程练习和推箱子游戏设计任务，评估学生的实践能力和问题解决能力。

考试评估：通过期末考试，评估学生对推箱子游戏算法和编程技巧的掌握程度。

六、教学安排本课程的教学安排将按照推箱子游戏的基本概念、算法和编程技巧的顺序进行，确保学生在有限的时间内掌握课程内容。

推箱子游戏课程设计c

推箱子游戏课程设计c一、课程目标知识目标：1. 学生能够理解推箱子游戏中的程序设计基本概念，掌握游戏规则逻辑表达方法。

2. 学生能够运用所学编程知识，设计并实现具有基本功能的推箱子游戏。

3. 学生理解并掌握游戏中坐标系统、角色移动算法和碰撞检测的相关知识。

技能目标：1. 学生通过实践操作，提高逻辑思维和问题解决能力，能够将复杂问题分解为简单步骤。

2. 学生培养独立编程能力，学会使用合适的编程语句和结构来实现游戏逻辑。

3. 学生通过小组合作，锻炼团队协作和沟通技巧，共同完成游戏设计任务。

情感态度价值观目标：1. 学生培养对计算机编程的兴趣，激发创造力和探究精神，形成积极的学习态度。

2. 学生在游戏设计过程中，学会面对挑战，培养坚持不懈和解决问题的耐心。

3. 学生通过推箱子游戏，理解编程在生活中的应用，认识到科技与日常生活紧密相关，增强社会责任感。

课程性质：本课程为信息技术课程，以实践操作为主，理论讲解为辅，注重培养学生的动手能力和实际应用能力。

学生特点：考虑到学生所在年级，课程内容设计将结合学生好奇心强、动手欲望高、团队协作意识逐步形成等特点。

教学要求：课程要求学生在掌握编程基础知识上，通过实践加深理解，鼓励创新思维和团队合作，实现游戏设计的目标。

教学过程将目标分解，确保每个学生都能达到具体的学习成果，为后续学习奠定基础。

二、教学内容1. 理论知识：- 游戏设计基本概念：游戏规则、游戏流程、角色与道具。

- 编程语言基础：变量、数据类型、运算符、控制结构（条件语句、循环语句）。

- 坐标系统与图形用户界面：理解坐标系在游戏中的应用，掌握简单图形界面的构建。

2. 实践操作：- 推箱子游戏规则设计：学习如何将游戏规则转化为程序逻辑。

- 角色移动算法：掌握角色在不同方向上的移动算法实现。

- 碰撞检测：学习如何检测角色与箱子、墙壁之间的碰撞。

3. 教学大纲：- 第一阶段：游戏设计基本概念学习，介绍推箱子游戏，引导学生分析游戏规则。

推箱子游戏实践报告

推箱子游戏设计实践报告一、游戏介绍在一定的范围里，要求人把木箱从开始位置推放到指定的位置。

由于箱子只能推不能拉，所以需要巧妙的利用有限的空间和通道，合理安排移动的次序和位置，才能顺利的完成任务。

该游戏一共有三道关卡，按下F1和F2可以切换关卡，并且通过移动上下左右键来控制小人的移动，从而推动箱子。

当人把箱子全部推到指定的目的地时过关。

附图：二、设计思想程序关键的地方有两部分，一是加载地图，二是用方向键控人的移动。

画地图主要由两个函数来实现，首先用加载位图函数来画出单个图片，然后再用调用加载位图的函数来画出整体的地图。

用方向键控制人的移动时，要注意全面地考虑到各方面的问题，需要详细的分析各种情况。

三、游戏的部分实现（1）加载地图部分程序void drawmap(HWND hWnd,int map[][8])//画地图函数{ HDC hdc;hdc = GetDC(hWnd);RECT rt;GetClientRect(hWnd, &rt);for(int y=0; y<8; y++){ for(int x=0; x<8; x++){switch(mapIndex[y][x]){case 2:drawImage(hdc, "aim.bmp", x * 48,y* 48,48,48);break;case 1:drawImage(hdc,"ground.bmp", x * 48,y* 48,48,48);break;case 0:drawImage(hdc, "wall.bmp", x* 48,y* 48,48,48);break;case 4:drawImage(hdc, "man.bmp", x* 48,y* 48,48,48);break;case 6:case 3:drawImage(hdc, "box.bmp", x* 48,y* 48,48,48);break;}}}ReleaseDC(hWnd, hdc)}voiddrawImage( HDC hdc, char * fileName, int x, int y , int w, int h){HDC srcdc = CreateCompatibleDC(hdc);HBITMAP hbmp = (HBITMAP)LoadImage(NULL, fileName,IMAGE_BITMAP, 0,0, LR_LOADFROMFILE);SelectObject(srcdc, hbmp);BitBlt(hdc, x,y, w,h, srcdc, 0,0, SRCCOPY );DeleteObject(hbmp);DeleteDC(srcdc);}（2）方向键移动部分程序voidkeydown_down(HWND hWnd){switch (mapIndex[humans.y+1][humans.x]){case 0:break;case 1:case 2:humans.y+=1;break;case 3://人的下面为箱子switch (mapIndex[humans.y+2][humans.x]){case 0:case 6:case 3:break;case 1:MoveResult();mapIndex[humans.y+2][humans.x]=3;mapIndex[humans.y+1][humans.x]=1;humans.y+=1;break;case 2:MoveResult();mapIndex[humans.y+2][humans.x]=6;mapIndex[humans.y+1][humans.x]=1;humans.y+=1;break;}break;case 6://箱子在目标上switch (mapIndex[humans.y+2][humans.x]){case 0:break;case 6:break;case 3:break;case 1:MoveResult();mapIndex[humans.y+2][humans.x]=3;mapIndex[humans.y+1][humans.x]=2;humans.y+=1;break;case 2:MoveResult();mapIndex[humans.y+2][humans.x]=6;mapIndex[humans.y+1][humans.x]=2;humans.y+=1;break;}break;}}voidMoveResult(){if(mapIndex[humans.y][humans.x]!=2) {mapIndex[humans.y][humans.x]=1;}else if(mapIndex[humans.y][humans.x]==2){mapIndex[humans.y][humans.x]=2;} }四、测试结果五、设计心得体会在设计这个程序中我主要学会了如何运用以下有关C语言的知识1)函数定义是要做到顾名思义是很重要的，它对读程序的人正确认识程序十分重要，在修改这个程序的过程中也能很快找到程序各模块的作用，大大增加了程序的可读性。

推箱子C课程设计报告

程序设计课程设计报告书学生：孟祥煜指导老师: 曾宇容课程设计名称：推箱子小游戏班级: 软件11４01一、设计任务介绍推箱子游戏得实现方法，并逐步介绍Ｃ语言图形编程得方法与技巧。

在C语言编写软件ＷIN-ＴC上，涉及软中断、二维数组、键盘操作及图形化函数等方法,显示器中断寄存器得设置、二维数组得定义、键盘上键值得获取、图形方式下光标得显示与定位,以及部分图形函数得使用实现了一个完整得推箱子游戏界面，界面清晰可见。

二、设计目得训练学生掌握C得数据结构与综合编程能力与游戏开发技巧等应用能力。

初步设想推箱子就是一款经典得益智小游戏,目得就是在训练人得逻辑思考能力。

在一个狭小得空间中，要求把木箱从起始位置处推放到指定位置。

在空间中有地方限制,稍不小心就会出现箱子无法移动或者通道被堵住得情况，而且箱子只能推不能拉,所以需要玩家巧妙得利用有限得空间与通道,合理得移动次序与位置,才能完成任务。

本游戏一共分1４关，由易到难,每一关都随着关数得增加来增加难度,从而达到吸引游戏玩家得目得。

三功能1、菜单功能,玩家通过瞧目录按对应得键可以进入对应得功能菜单,其中1进入游戏,2选择游戏关卡,3游戏帮助指南 ,4退出游戏。

2 游戏功能: 玩家可以通过操纵键盘中得上下左右键进入游戏进行各种操作来完成游戏。

3 循环功能:当玩家正在游戏得时候如果某一步走错了,导致游戏无法完成,可以随时重新开始游戏来完成关卡,游戏过程中可以随时按Ctrl＋V退出游戏。

4 顺序功能:当玩家完成了某一关之后，随着进入下一关门可以按顺序进入下一关,继续开始更有挑战得关卡,如果该关就是最后一关,则显示通关完成。

三、总体设计1、功能模块设计函数功能描述本程序包括5个模块，分别就是初始化模块、画图模块、移动箱子模块、移动小人模块与功能控制模块。

各个模块功能得功能描述如下：（1)初始化模块。

该模块包括屏幕初始化与游戏第一关到第最后一关得初始化。

屏幕初始化用于输出欢迎信息与操作提示,游戏每一关得初始化就是构建每一关得关卡。

c语音课程设计推箱子

c语音课程设计推箱子一、教学目标本课程的教学目标是使学生掌握C语音的基本语法和编程技巧，能够运用C语言进行简单的程序设计。

具体分为以下三个方面的目标：1.知识目标：使学生了解C语言的基本语法、数据类型、运算符、控制结构、函数等基本概念，掌握基本的编程技巧。

2.技能目标：通过课堂学习和上机实践，使学生能够熟练使用C语言进行编程，具备一定的程序设计能力。

3.情感态度价值观目标：培养学生对计算机科学的兴趣和热情，提高学生的问题解决能力和创新意识，培养学生的团队协作精神。

二、教学内容本课程的教学内容主要包括C语言的基本语法、数据类型、运算符、控制结构、函数等，具体包括以下几个方面的内容：1.C语言的基本语法和编程规范。

2.数据类型、运算符和表达式。

3.控制结构：条件语句、循环语句。

4.函数：函数的定义、调用和返回值。

5.数组、字符串和指针。

6.结构体、联合体和枚举类型。

7.文件操作。

三、教学方法为了达到本课程的教学目标，我们将采用以下几种教学方法：1.讲授法：通过讲解C语言的基本概念和语法，使学生掌握C语言的基本知识。

2.案例分析法：通过分析典型的C语言程序，使学生了解C语言的应用和编程技巧。

3.实验法：通过上机实验，使学生巩固所学的知识，提高编程能力。

4.讨论法：学生进行课堂讨论，培养学生的团队协作能力和创新意识。

四、教学资源为了保证本课程的顺利进行，我们将准备以下教学资源：1.教材：《C语言程序设计》。

2.参考书：提供一些相关的参考书籍，供学生课后自学。

3.多媒体资料：制作课件和教学视频，丰富教学手段。

4.实验设备：提供计算机实验室，供学生上机实践。

五、教学评估为了全面、客观地评估学生的学习成果，本课程将采用以下几种评估方式：1.平时表现：通过课堂参与、提问、讨论等环节，评估学生的学习态度和课堂表现。

2.作业：布置适量的作业，评估学生的编程能力和对知识的理解程度。

3.考试：包括期中考试和期末考试，以闭卷形式进行，全面考察学生的知识掌握和编程能力。

C语言推箱子课程设计(附报告)

（注：VC运行时，分别新建两个文件BotIt.cpp和Maps.h，再在同一目录下运行其中cpp）代码部分文件“BoxIt.cpp”如下：#include <stdio.h>#include <conio.h>#include <stdlib.h>#include "Maps.h"#define AISLE "" //走廊#define N_AISLE 0#define W ALL "■" //墙壁#define N_W ALL 1#define AIM "※" //未放箱子的目标#define N_AIM 2#define BOX "□" //未在目标位置的箱子#define N_BOX 3#define MAN "♀" //人#define N_MAN 4#define BOX_OK "▢" //■已经在目标位置的箱子[即已放箱子的目标]#define N_BOX_OK 5#define MAN_IN "* "#define N_MAN_IN 6#define ESC 0x1B#define UP 'w'#define DOWN 's'#define LEFT 'a'#define RIGHT 'd'#define RESTART 'r'#define SA VE 'x'#define PF(a) printf("%s",a)struct Point{int x;int y;}man_p;/*小人坐标*///struct Filename{int num;char name[20];}filename[10];//int N;/*存档总人数*/int M;/*关数*/int step;/*步数*/char x;void welcome();void ingame();void playgame(int map[14][16]);/*玩游戏*/void displaymap(int map[14][16]);/*展示当前游戏界面*/void findman(int &a,int &b,int map[14][16]);/*找到小人坐标位置*/unsigned char getKey();/*获得有效按键返回按键值*/int getmap(Point p,int map[14][16]);/*获得点p的地图值*/void moveman(Point man_p,Point manf_p,int X,int Y,int map[14][16]);/*人移动*/ void movebox(Point box_p,Point boxf_p,int map[14][16]);/*移箱子*/int win(int map[14][16]);/*判断当无※时赢*/void choosemap(int map[14][16],int i);/*选关*/void changemap(int map[14][16],int mapx[14][16]);/*换地图*/void save();/*保存当前*/int read();/*读取存档*/int a=0,b=0,c=0,d=0;//system("color a");int main(){x='0';while(x!='4'){welcome();if(x=='3'&&read()==0) continue;if(x!='4') ingame();}return 0;}void displaymap(int map[14][16]){int i,j;system("cls");printf("\t\t\tMisson %d\n\n",M);for(i=0;i<14;i++){for(j=0;j<16;j++){ switch(map[i][j]){case N_AISLE:PF(" ");break;case N_W ALL:PF(W ALL);break;case N_AIM:PF(AIM);break;case N_BOX:PF(BOX);break;case N_MAN:PF(MAN);break;case N_BOX_OK:PF(BOX_OK);break;case N_MAN_IN:PF(MAN_IN);default:;}}if(i==2) printf("Press the key:");if(i==3) printf("'w'、'a'、's'、'd' to control ♀");if(i==5) printf("'r' to replay");if(i==7) printf("'x' to save");if(i==9) printf("'ESC' back to the menu");if(i==12) printf("Step:%d",step);PF("\n");}}void playgame(int map[14][16]){ int X,Y;Point manf_p,manff_p;//findman(man_p.x,man_p.y,map);displaymap(map);step=0;while(win(map)==0){findman(man_p.x,man_p.y,map);//printf("输入：");switch(getKey()){case DOWN://printf("\n下\n");manf_p.x=man_p.x;manff_p.x=man_p.x;manf_p.y=man_p.y+1,manff_p.y=man_p.y+2;X=0,Y=1;break;case UP://printf("\n上\n");manf_p.x=man_p.x;manff_p.x=man_p.x;manf_p.y=man_p.y-1,manff_p.y=man_p.y-2;X=0,Y=-1;break;case LEFT://printf("\n左\n");manf_p.x=man_p.x-1;manff_p.x=man_p.x-2;manf_p.y=man_p.y;manff_p.y=man_p.y;X=-1,Y=0;break;case RIGHT://printf("\n右\n");manf_p.x=man_p.x+1;manff_p.x=man_p.x+2;manf_p.y=man_p.y;manff_p.y=man_p.y;X=1,Y=0;break;case SA VE:save();continue;case RESTART:changemap(map,map0);M--;continue;case ESC:system("cls");changemap(map,map0);M=44;continue;default:;}//printf("\n人位置[%d %d]\n",man_p.y,man_p.x);//printf("\n人前方[%d %d]人前前方[%d %d]\n",manf_p.y,manf_p.x,manff_p.y,manff_p.x);switch(getmap(manf_p,map)){case N_AISLE://0case N_AIM://2moveman(man_p,manf_p,X,Y,map);break;case N_BOX://3case N_BOX_OK://5if(getmap(manff_p,map)==N_AISLE||getmap(manff_p,map)==N_AIM){movebox(manf_p,manff_p,map);moveman(man_p,manf_p,X,Y,map);}break;default:;}displaymap(map);}if(M==42) printf("\n\tCongratulation! Pass All!!!\n\n\n\t");}void findman(int &a,int &b,int map[14][16]){int i,j;for(i=0;i<14;i++)for(j=0;j<16;j++)if(map[i][j]==N_MAN||map[i][j]==N_MAN_IN)a=j,b=i;}unsigned char getKey(){unsigned char key;do{key = getch();}while(key != RESTART && key != ESC && key != LEFT && key != RIGHT && key != UP && key != DOWN && key!='x'&& key!=0x1B &&key!='r');return key;}int getmap(Point p,int map[14][16]){//printf("\n前方值%d\n",map[p.y][p.x]);return map[p.y][p.x];}void moveman(Point man_p,Point manf_p,int X,int Y,int map[14][16]){map[manf_p.y][manf_p.x]=(map[manf_p.y][manf_p.x]==N_AIM||map[manf_p.y][manf_p.x]==N_BOX _OK)?N_MAN_IN:N_MAN;map[man_p.y][man_p.x]=map[man_p.y][man_p.x]==N_MAN_IN?N_AIM:N_AISLE; //printf("\n移动后—小人位置值%d\n",map[manf_p.y][manf_p.x]);//printf("\n移动后—原小人位置值%d\n",map[man_p.y][man_p.x]);man_p.x+=X,man_p.y+=Y;step++;}void movebox(Point box_p,Point boxf_p,int map[14][16]){map[boxf_p.y][boxf_p.x]=map[boxf_p.y][boxf_p.x]==N_AISLE?N_BOX:N_BOX_OK; map[box_p.y][box_p.x]=map[box_p.y][box_p.x]==N_BOX?N_AISLE:N_AIM;}int win(int map[14][16]){int i,j;for(i=0;i<14;i++)for(j=0;j<16;j++)if(map[i][j]==2||map[i][j]==6)return 0;return 1;}void choosemap(int map[14][16],int i){switch(i){case 1:changemap(map,map1);break;case 2:changemap(map,map2);break;case 3:changemap(map,map3);break;case 4:changemap(map,map4);break;case 5:changemap(map,map5);break;case 6:changemap(map,map6);break;case 7:changemap(map,map7);break;case 8:changemap(map,map8);break;case 9:changemap(map,map9);break;case 10:changemap(map,map10);break;case 11:changemap(map,map11);break;case 12:changemap(map,map12);break;case 13:changemap(map,map13);break;case 14:changemap(map,map14);break;case 15:changemap(map,map15);break;case 16:changemap(map,map16);break;case 17:changemap(map,map17);break;case 18:changemap(map,map18);break;case 19:changemap(map,map19);break;case 20:changemap(map,map20);break;case 21:changemap(map,map21);break;case 22:changemap(map,map22);break;case 23:changemap(map,map23);break;case 24:changemap(map,map24);break;case 25:changemap(map,map25);break;case 26:changemap(map,map26);break;case 27:changemap(map,map27);break;case 28:changemap(map,map28);break;case 29:changemap(map,map29);break;case 30:changemap(map,map30);break;case 31:changemap(map,map31);break;case 32:changemap(map,map32);break;case 33:changemap(map,map33);break;case 34:changemap(map,map34);break;case 35:changemap(map,map35);break;case 36:changemap(map,map36);break;case 37:changemap(map,map37);break;case 38:changemap(map,map38);break;case 39:changemap(map,map39);break;case 40:changemap(map,map40);break;case 41:changemap(map,map41);break;case 42:changemap(map,map42);break;default:;}}void changemap(int map[14][16],int mapx[14][16]) {int i,j;for(i=0;i<14;i++)for(j=0;j<16;j++)map[i][j]=mapx[i][j];}void save(){int i,j;FILE *in;/*char filename[10][20];printf("put in the filename:");scanf("%s",filename);*/if((in=fopen("save game","w"))==NULL){printf("cannot open this file\n");exit(0);}for(i=0;i<14;i++)for(j=0;j<16;j++)fwrite(&map[i][j],sizeof(int),1,in);fwrite(&M,sizeof(int),1,in);fwrite(&step,sizeof(int),1,in);printf("\nGame Misson %d saved!\n",M);system("pause");fclose(in);}int read(){int i,j;FILE *out;/*char filename[10][20];printf("put in the filename:");scanf("%s",filename);*/if((out=fopen("save game","r"))==NULL){printf("\ncannot open this file\n");return 0;}else{for(i=0;i<14;i++)for(j=0;j<16;j++)fread(&map[i][j],sizeof(int),1,out);fread(&M,sizeof(int),1,out);fread(&step,sizeof(int),1,out);printf("\nM=%d\n",M);printf("Game read!\n");fclose(out);return 1;}}void welcome(){printf("\t\t ♀Box It \n");printf("\t\t1:New Game.\n\t\t2:Select Misson.\n\t\t3:Read Record.\n\t\t4:Exit Game.\n Choose-- ");switch(x=getch()){case '1':printf("1:");M=1;choosemap(map,M);break;case '2':printf("2:");printf("1-42 Misson to Choose:");scanf("%d",&M);choosemap(map,M);break;case '3':printf("3:");if(read()==1) printf("\n%d misson opened\n",M);break;case '4':printf("4:Exit Game.\n\n"); break;default:printf("Not 1-4\n");}}void ingame(){while(M<43){playgame(map);step=0;M++;choosemap(map,M);}}文件“Maps.h”如下：int map[14][16];int map0[14][16]={{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},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{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},{0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0},{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},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {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}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map1[14][16]={{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}, {0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0}, {0,0,0,0,0,1,2,1,0,0,0,0,0,0,0,0}, {0,0,0,0,0,1,0,1,1,1,1,0,0,0,0,0}, {0,0,0,1,1,1,3,0,3,2,1,0,0,0,0,0}, {0,0,0,1,2,0,3,4,1,1,1,0,0,0,0,0}, {0,0,0,1,1,1,1,3,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,1,2,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0}, {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}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map2[14][16]={{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0}, {0,0,0,1,4,0,0,1,0,0,0,0,0,0,0,0}, {0,0,0,1,0,3,3,1,0,1,1,1,0,0,0,0}, {0,0,0,1,0,3,0,1,0,1,2,1,0,0,0,0}, {0,0,0,1,1,1,0,1,1,1,2,1,0,0,0,0}, {0,0,0,0,1,1,0,0,0,0,2,1,0,0,0,0}, {0,0,0,0,1,0,0,0,1,0,0,1,0,0,0,0}, {0,0,0,0,1,0,0,0,1,1,1,1,0,0,0,0}, {0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {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}, };int map3[14][16]={{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}, {0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0}, {0,0,0,0,1,0,0,0,0,0,1,1,1,0,0,0}, {0,0,0,1,1,3,1,1,1,0,0,0,1,0,0,0},{0,0,0,1,0,2,2,1,0,3,0,1,1,0,0,0}, {0,0,0,1,1,2,2,1,0,0,0,1,0,0,0,0}, {0,0,0,0,1,1,1,1,1,1,1,1,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {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}, };int map4[14][16]={{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}, {0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0}, {0,0,0,1,1,0,0,1,0,0,0,0,0,0,0,0}, {0,0,0,1,4,3,0,1,0,0,0,0,0,0,0,0}, {0,0,0,1,1,3,0,1,1,0,0,0,0,0,0,0}, {0,0,0,1,1,0,3,0,1,0,0,0,0,0,0,0}, {0,0,0,1,2,3,0,0,1,0,0,0,0,0,0,0}, {0,0,0,1,2,2,5,2,1,0,0,0,0,0,0,0}, {0,0,0,1,1,1,1,1,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map5[14][16]={{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}, {0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0}, {0,0,0,0,1,4,0,1,1,1,0,0,0,0,0,0}, {0,0,0,0,1,0,3,0,0,1,0,0,0,0,0,0}, {0,0,0,1,1,1,0,1,0,1,1,0,0,0,0,0}, {0,0,0,1,2,1,0,1,0,0,1,0,0,0,0,0}, {0,0,0,1,2,3,0,0,1,0,1,0,0,0,0,0}, {0,0,0,1,2,0,0,0,3,0,1,0,0,0,0,0}, {0,0,0,1,1,1,1,1,1,1,1,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {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}, };int map6[14][16]={{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0},{0,1,0,0,0,2,1,1,1,0,1,0,0,0,0,0}, {0,1,0,1,0,1,0,0,0,0,1,1,0,0,0,0}, {0,1,0,1,0,3,0,3,1,2,0,1,0,0,0,0}, {0,1,0,1,0,0,5,0,0,1,0,1,0,0,0,0}, {0,1,0,2,1,3,0,3,0,1,0,1,0,0,0,0}, {0,1,1,0,0,0,0,1,0,1,0,1,1,1,0,0}, {0,0,1,0,1,1,1,2,0,0,0,0,4,1,0,0}, {0,0,1,0,0,0,0,0,1,1,0,0,0,1,0,0}, {0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map7[14][16]={{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}, {0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0}, {0,0,0,0,0,0,1,1,0,0,1,0,4,1,0,0}, {0,0,0,0,0,0,1,0,0,0,1,0,0,1,0,0}, {0,0,0,0,0,0,1,3,0,3,0,3,0,1,0,0}, {0,0,0,0,0,0,1,0,3,1,1,0,0,1,0,0},{0,0,0,0,1,2,2,2,2,2,0,0,1,1,0,0}, {0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map8[14][16]={{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}, {0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0}, {0,0,0,0,1,1,1,0,0,0,0,1,0,0,0,0}, {0,0,0,1,1,2,0,3,1,1,0,1,1,0,0,0}, {0,0,0,1,2,2,3,0,3,0,0,4,1,0,0,0}, {0,0,0,1,2,2,0,3,0,3,0,1,1,0,0,0}, {0,0,0,1,1,1,1,1,1,0,0,1,0,0,0,0}, {0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0}, {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}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},};int map9[14][16]={{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}, {0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0}, {0,0,0,1,0,0,1,1,0,0,0,1,0,0,0,0}, {0,0,0,1,0,0,0,3,0,0,0,1,0,0,0,0}, {0,0,0,1,3,0,1,1,1,0,3,1,0,0,0,0}, {0,0,0,1,0,1,2,2,2,1,0,1,0,0,0,0}, {0,0,1,1,0,1,2,2,2,1,0,1,1,0,0,0}, {0,0,1,0,3,0,0,3,0,0,3,0,1,0,0,0}, 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{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}, };int map32[14][16]={{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}, {0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,0},{0,0,0,0,0,1,2,0,3,2,1,0,0,0,0,0}, {0,0,0,0,1,1,1,0,0,3,1,1,0,0,0,0}, {0,0,0,0,1,0,3,0,0,3,0,1,0,0,0,0}, {0,0,0,0,1,0,1,3,1,1,0,1,0,0,0,0}, {0,0,0,0,1,0,0,0,4,0,0,1,0,0,0,0}, {0,0,0,0,1,1,1,1,1,1,1,1,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {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}, };int map33[14][16]={{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0}, {0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0}, {0,0,1,0,0,0,1,3,0,0,0,1,1,1,0,0}, {0,0,1,0,0,0,3,0,0,0,3,3,0,1,0,0}, {0,0,1,0,3,3,0,1,3,0,0,0,0,1,0,0}, {0,0,1,1,0,0,0,3,0,0,0,3,0,1,0,0}, {1,1,1,1,1,1,0,1,3,1,1,1,1,1,0,0}, {1,2,2,4,0,1,3,0,0,1,0,0,0,0,0,0}, {1,2,1,2,2,0,0,3,1,1,0,0,0,0,0,0},{1,2,2,2,2,0,0,0,1,0,0,0,0,0,0,0}, {1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map34[14][16]={{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0}, {1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0}, {1,0,3,0,1,3,0,1,0,3,1,1,3,0,1,0}, {1,0,1,0,0,3,0,1,0,0,0,0,0,0,1,0}, {1,0,0,0,1,1,3,1,3,1,1,3,3,0,1,0}, {1,0,1,0,1,0,2,2,2,0,1,0,0,0,1,0}, {1,0,3,0,0,2,0,1,0,2,3,0,1,0,1,0}, {1,0,3,1,4,3,2,2,2,1,0,1,0,0,1,0}, {1,0,0,0,0,2,0,1,0,2,0,0,3,0,1,0}, {1,0,1,1,2,3,1,1,1,3,2,0,1,0,1,0}, {1,0,1,0,3,2,2,2,2,2,0,1,1,0,1,0}, {1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0}, {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map35[14][16]={{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0}, {0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,0}, {0,0,0,1,0,1,0,3,0,3,0,1,0,0,0,0}, {0,0,0,1,0,0,5,2,1,0,0,1,0,0,0,0}, {0,0,0,1,1,0,1,2,4,2,1,1,0,0,0,0}, {0,0,0,1,1,3,1,1,1,5,1,1,1,0,0,0}, {0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0}, {0,0,0,1,0,0,0,1,1,0,1,0,1,0,0,0}, {0,0,0,1,1,1,1,1,1,0,0,0,1,0,0,0}, {0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, };int map36[14][16]={{1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0}, {1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0}, {1,0,3,3,0,0,0,1,1,1,0,0,0,0,0,0}, {1,0,0,3,0,3,3,3,0,1,1,1,1,1,0,0}, {1,1,0,1,1,0,2,2,2,0,0,0,0,1,1,0}, {0,1,0,1,4,1,2,2,2,1,1,1,3,0,1,0},{1,1,0,1,0,3,2,2,2,3,0,1,0,1,1,0}, {1,0,0,1,1,1,1,1,0,1,1,1,0,1,0,0}, {1,0,0,0,0,0,0,3,0,0,0,3,0,1,0,0}, {1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0}, {0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0}, {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}, };int map37[14][16]={{0,1,1,1,1,0,0,0,0,0,0,1,1,1,1,0}, {0,1,2,2,1,0,0,0,0,0,0,1,2,2,1,0}, {0,1,2,2,1,1,1,1,1,1,1,1,2,2,1,0}, {0,1,0,3,0,2,2,2,2,2,2,3,2,3,1,0}, {0,1,3,0,3,0,3,0,3,0,3,0,3,0,1,0}, {0,1,0,3,0,3,0,3,4,3,0,3,0,3,1,0}, {0,1,3,0,3,0,3,0,3,0,3,0,3,0,1,0}, {0,1,0,3,0,3,0,3,0,3,0,3,0,3,1,0}, {0,1,3,2,3,2,2,2,2,2,2,0,3,0,1,0}, {0,1,2,2,1,1,1,1,1,1,1,1,2,2,1,0}, {0,1,2,2,1,0,0,0,0,0,0,1,2,2,1,0}, {0,1,1,1,1,0,0,0,0,0,0,1,1,1,1,0},。

C推箱子课程设计报告

C推箱子课程设计报告The document was prepared on January 2, 2021目：基于学武汉理工大学华夏学院课程设计报告书课程名称：软件综合系名：信息工程系专业班级: 姓名：指导教师：2013年元月18 H 课程设计任务书学生姓名： ________________ 专业班级：软件2101指导教师： ________ 工作单位：信息系计算机教研室设计题目：基于VC++的“推箱子”游戏软件的开发与测试初始条件：1. 已完成c 语言程序设计、软件工程、数据库概论或数据库技术等课程的学习2. 系统环境：Intel 奔腾4以上处理器，1G 以上内存，40G 以上硬盘，Windows 编程环境，Visual O+编程工具要求完成白勺主要任务：（在规定的时间内完成下列任务）（包括课程设计工作量及其技术要求，以及说明书撰写等具体要求）主要任务：系统完成用户登录及管理员增加、删除、修改、查询功能1. 按系统功能进行游戏背景区域（移动区域、围墙）的设计•与绘制:2. 根据系统实现的功能，绘出功能图、系统流程图3. 模块调试，运行结果正确，并经教师检查及答辩；设计报告撰写格式要求：（课程设计报告按统一通用格式书写）具体内容如下：① 设计任务与要求② 总体方案与说明③程序主要模块的流程图④问题分析与解决方案的设计思想和改进设想⑤系统的调试记录、调试报告（即在调试中遇到的主要问题、解决方法及改进设想）⑥小结与体会附录：①使用说明②参考资料时间安排：2012年12月29日一-2013年元18日（共三周）指导教师签字：2012年12月28日系主任签字：年月日1引言最经典的推箱子游戏，类似的游戏你一定早就玩过。

要控制搬运工上下左右移动，来将箱子推到指定地点记得儿年前，《推箱子》在机上刮起了一股不小的益智类游戏的旋风，现在许多资深玩家也都对《推箱子》赞不绝口，可见有深度的益智类游戏是非常受大家欢迎的。

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程序设计课程设计报告书欧阳光明（2021.03.07）学生：孟祥煜指导老师：曾宇容课程设计名称：推箱子小游戏班级：软件11401一、设计任务介绍推箱子游戏的实现方法，并逐步介绍C语言图形编程的方法和技巧。

在C语言编写软件WIN-TC上，涉及软中断、二维数组、键盘操作及图形化函数等方法，显示器中断寄存器的设置、二维数组的定义、键盘上键值的获取、图形方式下光标的显示和定位，以及部分图形函数的使用实现了一个完整的推箱子游戏界面，界面清晰可见。

二、设计目的训练学生掌握C的数据结构和综合编程能力和游戏开发技巧等应用能力。

初步设想推箱子是一款经典的益智小游戏，目的是在训练人的逻辑思考能力。

在一个狭小的空间中，要求把木箱从起始位置处推放到指定位置。

在空间中有地方限制，稍不小心就会出现箱子无法移动或者通道被堵住的情况，而且箱子只能推不能拉，所以需要玩家巧妙的利用有限的空间和通道，合理的移动次序和位置，才能完成任务。

本游戏一共分14关，由易到难，每一关都随着关数的增加来增加难度，从而达到吸引游戏玩家的目的。

三功能1、菜单功能，玩家通过看目录按对应的键可以进入对应的功能菜单，其中1进入游戏，2选择游戏关卡，3游戏帮助指南，4退出游戏。

2 游戏功能：玩家可以通过操纵键盘中的上下左右键进入游戏进行各种操作来完成游戏。

3 循环功能：当玩家正在游戏的时候如果某一步走错了，导致游戏无法完成，可以随时重新开始游戏来完成关卡，游戏过程中可以随时按Ctrl+V退出游戏。

4顺序功能：当玩家完成了某一关之后，随着进入下一关门可以按顺序进入下一关，继续开始更有挑战的关卡，如果该关是最后一关，则显示通关完成。

三、总体设计1、功能模块设计函数功能描述本程序包括5个模块，分别是初始化模块、画图模块、移动箱子模块、移动小人模块和功能控制模块。

各个模块功能的功能描述如下：（1）初始化模块。

该模块包括屏幕初始化和游戏第一关到第最后一关的初始化。

屏幕初始化用于输出欢迎信息和操作提示，游戏每一关的初始化是构建每一关的关卡。

（2）画图模块。

该模块主要是被其他模块调用，用于画墙、在空地画箱子、在目的地画箱子、画小人和画目的地。

（3）移动箱子模块。

该模块用于移动箱子，包括目的地之间、空地之间和目的地与空地之间箱子的移动。

（4）移动小人模块。

该模块是几个功能函数的集合，包括屏幕输出功能、指定位置状态判断功能和关卡重置功能。

（5）功能控制模块。

该模块是几个功能函数的集合，包括屏幕输出功能、指定位置状态判断功能和关卡重置功能。

任务执行流程图数据流程图游戏从第一关开始，按上下左右方向键控制小人移动来推动箱子，可以在游戏中的任何时候按Esc键退出。

如果游戏无成功希望，可以按CTRI+V回到当前任务的开始状态；如果成功完成当前关，则进入下一关，如果当前关是最后一关，则显示通关信息，提示游戏结束。

小人移动流程图小人移动的方向有4个，move（）函数（处理小人移动的函数）对这4个方向移动的处理都一致，只是调用函数时的参数有所不同。

首先判断小人移动的方向，然后根据小人所处世的当前状态、下一步状态或者下下一步状态进行适当的处理。

2、数据结构设计设置全局变量定义int v[13][17]，用于记录箱子在屏幕中哪个坐标上。

其中“0”表示什么都没有。

定义了int num_guan指现在的关数，int num_xiang;箱子的个数，int move_num人行走的次数；int push_num人推箱子的次数；定义数组int M[2]表示人所在的位置; M[0] 是人所在位置的横坐标M[1]是人所在位置的纵坐标；定义数组int door[2]表示门所在的坐标，door[0]是“下一关门”的横坐标，door[1]是“下一关门”的纵坐标。

3主函数下面让我们来看看主函数的操作流程，主函数首先进入游戏目录，提醒游戏玩家输入对应的数字来进行的对应的操作，如果玩家输入1则进入游戏第一关，输入2选择关卡进入，输入3显示游戏帮助菜单，帮助玩家更好的理解这个游戏的规则，输入4退出游戏。

进入游戏后，首先调用chushi1()函数初始化要使用的页面，之后程序进入play()函数，play()函数中通过if语句通过判断i的值来初始化不同的关卡（设置好关卡的人，箱子，箱子的目的地，门），接着程序进入control函数通过获取玩家在键盘上的操作来分别对游戏人物进行操作，最后通过wancheng函数来对操作进行判断，如果判断成功的话则进入下一关游戏。

反之则失败。

一直通过到最后一个则通关成功。

四代码设计（或编程）①单个模块或函数处理将整个系统程序分解为各个单独的模块，即C的函数，各模块可单独编译；②整个程序的联调在各个模块基本的单独编译通过后，用主函数将他们集成在一起，实现调用，并进行输出数据的可靠性、输入数据及系统的容错性测试。

五程序说明书功能模块1.play()2.control()3.Daying()4.Wancheng()解析1. play()play()函数在这个程序中的功能主要是按1进入游戏系统后自动按照顺序进行游戏和完成每一关游戏后进入下一关关卡，其中变量i控制关卡的关数，j控制进入下一关，首先利用if语句将i从一开始，初始化完成第一关，通过play()函数中的内嵌函数control()函数来控制游戏中人和箱子的各种操作，当过关了control()函数return 0.即j=control()中的j等于0，游戏进入下一关。

依次类推，当完成14关之后j=1游戏结束。

void play(){int i,j=0;for(i=1;i<=20;i++){tem=0;system("cls");//清屏if(i==1) chushihua_guan1(); //设置关卡1的墙，箱子，人。

门else if(i==2) chushihua_guan2();else if(i==3) chushihua_guan3();else if(i==4) chushihua_guan4();else if(i==5) chushihua_guan5();else if(i==6) chushihua_guan6();else if(i==7) chushihua_guan7();else if(i==8) chushihua_guan8();else if(i==9) chushihua_guan9();else if(i==10) chushihua_guan10();else if(i==11) chushihua_guan11();else if(i==12) chushihua_guan12();else if(i==13) chushihua_guan13();else if(i==14) chushihua_guan14();j=control();if(j==1){system("cls");return ;}}}2 control()Control()在这个程序中主要控制游戏中人物和箱子的运动，当玩家在键盘进行各种操作时系统会给出不同的反应。

模块中定义了command变量，来接收玩家在键盘上的各种操作，首先调用getch()函数为command赋一个初值command=（“空字符”）22, 进入界面图表初始化，根据num_guan的大小来具体初始化对应的关卡（箱子，人，墙的图标），接着模块进入了“控制阶段”，再次通过getch()函数command 获得一个字符，当command=”H”(也就是↑)时人进行向上前进的动作，对应的(“P””K””M”)进行人的↓、←、→操作，以及人的上下左右推箱子的动作，同时当人的前进的方向是墙或者双层箱子的时候人则不能运动，人推箱子如果箱子的前面是墙或者箱子的话箱子和人都不能运动。

运动过程中如果玩家输入ctrl+c时游戏此关卡重新开始，同时在control函数中调用一个win变量来判断是否完成了游戏，调用内嵌函数wancheng() 函数,当((win=wancheng())==1&&v[door[0]][door[1]]==ren)时，表示这个关卡已经完成了，return 0,即把play()函数中的j赋值等于0，游戏进入下一个关。

游戏过程玩家每操作一步，系统通过控制daying()函数来控制画面中目标的运动，来达到游戏的目的。

int control(){int win;int temm;char command; //存放接收到命令while(1){command=getch(); //输入一个字符/* if(command==1)if(command==22){if(num_guan==1){system("cls");chushihua_guan1();}else if(num_guan==2) {system("cls");chushihua_guan2();}else if(num_guan==3) {system("cls");chushihua_guan3();}else if(num_guan==4) {system("cls");chushihua_guan4();}else if(num_guan==5) {system("cls");chushihua_guan5();}else if(num_guan==6) {system("cls");chushihua_guan6();}else if(num_guan==7) {system("cls");chushihua_guan7();}else if(num_guan==8) {system("cls");chushihua_guan8();}else if(num_guan==9) {system("cls");chushihua_guan9();}else if(num_guan==10) {system("cls");chushihua_guan10();}else if(num_guan==11){system("cls");chushihua_guan11();}else if(num_guan==12) {system("cls");chushihua_guan12();}else if(num_guan==1113) {system("cls");chushihua_guan13();}else if(num_guan==14) {system("cls");chushihua_guan14();}}if(command==-32) //F11,F12:-123,-122{temm=0;command=getch();if(temm==0&&command=='H' && (v[M[0]-1][M[1]]==xiang||v[M[0]-1][M[1]]==oxiang) && v[M[0]-2][M[1]]==jia) //箱子在人的上边或者人的上边的箱子已经放好了且人的上面两行就是箱子的家{ v[M[0]-1][M[1]]=ren; //H指的是向上推箱子到家v[M[0]][M[1]]=0;v[M[0]-2][M[1]]=oxiang;M[0]=M[0]-1;move_num=move_num+1;push_num=push_num+1;temm=temm+1;}if(temm==0&&command=='P' && (v[M[0]+1][M[1]]==xiang||v[M[0]+1][M[1]]==oxiang) && v[M[0]+2][M[1]]==jia){ v[M[0]+1][M[1]]=ren;temm=temm+1; //P 指向下推箱子到家v[M[0]][M[1]]=0;v[M[0]+2][M[1]]=oxiang;M[0]=M[0]+1;move_num=move_num+1;push_num=push_num+ 1;}if(temm==0&&command=='K' && (v[M[0]][M[1]-1]==xiang||v[M[0]][M[1]-1]==oxiang) && v[M[0]][M[1]-2]==jia){ v[M[0]][M[1]-1]=ren; //K指向左推箱子到家v[M[0]][M[1]]=0;temm=temm+1;v[M[0]][M[1]-2]=oxiang;M[1]=M[1]-1;move_num=move_num+1;push_num=push_num+1;}if(temm==0&&command=='M' && (v[M[0]][M[1]+1]==xiang||v[M[0]][M[1]+1]==oxiang) &&v[M[0]][M[1]+2]==jia){ v[M[0]][M[1]+1]=ren;temm=temm+1; //M 指向右推箱子到家v[M[0]][M[1]]=0;v[M[0]][M[1]+2]=oxiang;M[1]=M[1]+1;move_num=move_num+1;push_num=push_num +1;}//H指人把箱子向上推，不是推到家if(temm==0&&command=='H' && (v[M[0]-1][M[1]]==xiang||v[M[0]-1][M[1]]==oxiang) && (v[M[0]-2][M[1]]==0||v[M[0]-2][M[1]]==jia)){ v[M[0]-1][M[1]]=ren;temm=temm+1;v[M[0]][M[1]]=0;v[M[0]-2][M[1]]=xiang;M[0]=M[0]-1;move_num=move_num+1;push_num=push_num+1;}if(temm==0&&command=='H' && (v[M[0]-1][M[1]]==0||v[M[0]-1][M[1]]==jia||v[M[0]-1][M[1]]==men)) { v[M[0]-1][M[1]]=ren;temm=temm+1;v[M[0]][M[1]]=0;move_num=mov e_num+1;M[0]=M[0]-1;} //人向上走if(temm==0&&command=='P' && (v[M[0]+1][M[1]]==0||v[M[0]+1][M[1]]==jia||v[M[0]+1][M[1]]==men)){ v[M[0]+1][M[1]]=ren;temm=temm+1; //人的下面为空或者下面为家或者下面为门的时候向下走v[M[0]][M[1]]=0;M[0]=M[0]+1;move_num=move_num+1;}if(temm==0&&command=='P' && (v[M[0]+1][M[1]]==xiang||v[M[0]+1][M[1]]==oxiang) && (v[M[0]+2][M[1]]==0||v[M[0]+2][M[1]]==jia)){ v[M[0]+1][M[1]]=ren; //人的下面为箱子或者下面已经有箱子找到家了切下面的下面是空或者下面的下面是家,人把箱子向下推，人往下面走v[M[0]][M[1]]=0;temm=temm+1;v[M[0]+2][M[1]]=xiang;M[0]=M[0]+1;move_num=move_num+1;push_num=push_num+ 1;}if(temm==0&&command=='K' && (v[M[0]][M[1]-1]==0||v[M[0]][M[1]-1]==jia||v[M[0]][M[1]-1]==men)){ v[M[0]][M[1]-1]=ren;temm=temm+1; //人的左边为空或者左边是家或者左边是门的时候人可以往左边走v[M[0]][M[1]]=0;M[1]=M[1]-1;move_num=move_num+1;}if(temm==0&&command=='K' && (v[M[0]][M[1]-1]==xiang||v[M[0]][M[1]-1]==oxiang) && (v[M[0]][M[1]-2]==0||v[M[0]][M[1]-2]==jia)){ v[M[0]][M[1]-1]=ren;temm=temm+1;//人的左边是箱子或者人的左边是已经找到家的箱子切人的左边的左边是空或者人的左边的左边是家的时候箱子可以往左边移动，v[M[0]][M[1]]=0;v[M[0]][M[1]-2]=xiang;M[1]=M[1]-1;move_num=move_num+1;push_num=push_num+1;}if(temm==0&&command=='M' && (v[M[0]][M[1]+1]==0||v[M[0]][M[1]+1]==jia||v[M[0]][M[1]+1]= =men)){v[M[0]][M[1]+1]=ren;temm=temm+1;v[M[0]][M[1]]=0;M[1]= M[1]+1;move_num=move_num+1;} //人的右边是空的或者人的右边是家或者人的右边的门的时候人可以往右边走。