求助，谁有一个有趣的C语言小程序，而且必须要有源代码！！
# include & ltstdio.h & gt
# include & ltconio.h & gt
# include & ltstdlib.h & gt
# include & ltwindows.h & gt
int m = 0；？//m代表哪一级。
结构映射{ short a[9][11]；};
结构映射map[5]={ 0，0，0，0，0，0，0，0，0，0，0，0，？//***5级，每级9行。
？0,1,1,1,1,1,1,1,0,0,0,
？0,1,0,0,0,0,0,1,1,1,0,
？1,1,4,1,1,1,0,0,0,1,0, ?//0空地，1墙
1,5,0,0,4,0,0,4,0,1,0, ?//4是盒子，5是人。
？1,0,3,3,1,0,4,0,1,1,0, ?//3是目的地
？1,1,3,3,1,0,0,0,1,0,0, ?//7表示盒子在目的地(4+3)
？0,1,1,1,1,1,1,1,1,0,0, ?//8是目的地的人(5+3)
？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,
？0,0,1,5,0,1,1,1,0,0,0,
？0,0,1,0,4,0,0,1,0,0,0,
0,1,1,1,0,1,0,1,1,0,0,
？0,1,3,1,0,1,0,0,1,0,0,
？0,1,3,4,0,0,1,0,1,0,0,
？0,1,3,0,0,0,4,0,1,0,0,
？0,1,1,1,1,1,1,1,1,0,0,
？0,0,0,0,0,0,0,0,0,0,0,
？0,0,0,1,1,1,1,1,1,1,0,
？0,0,1,1,0,0,1,0,5,1,0,
？0,0,1,0,0,0,1,0,0,1,0,
0,0,1,4,0,4,0,4,0,1,0,
？0,0,1,0,4,1,1,0,0,1,0,
？1,1,1,0,4,0,1,0,1,1,0,
？1,3,3,3,3,3,0,0,1,0,0,
？1,1,1,1,1,1,1,1,1,0,0,
？0,1,1,1,1,1,1,1,1,1,0,
？0,1,0,0,1,1,0,0,0,1,0,
？0,1,0,0,0,4,0,0,0,1,0,
？0,1,4,0,1,1,1,0,4,1,0,
0,1,0,1,3,3,3,1,0,1,0,
？1,1,0,1,3,3,3,1,0,1,1,
？1,0,4,0,0,4,0,0,4,0,1,
？1,0,0,0,0,0,1,0,5,0,1,
？1,1,1,1,1,1,1,1,1,1,1,
？0,0,0,0,0,0,0,0,0,0,0,
？0,0,0,1,1,1,1,1,1,0,0,
？0,1,1,1,0,0,0,0,1,0,0,
？1,1,3,0,4,1,1,0,1,1,0,
1,3,3,4,0,4,0,0,5,1,0,
？1,3,3,0,4,0,4,0,1,1,0,
？1,1,1,1,1,1,0,0,1,0,0,
？0,0,0,0,0,1,1,1,1,0,0,
？0,0,0,0,0,0,0,0,0,0,0 };
void DrMap()？//画一张地图
{控制台_光标_信息光标_信息={1，0 }；//隐藏光标设置
SetConsoleCursorInfo(GetStdHandle(STD _ OUTPUT _ HANDLE)，& ampcursor _ info)；
Printf("\n\n \t\t\b推框")；
printf(" \ n \ t ")；
for(int I = 0；我& lt9;i++)
{ for(int j = 0；j & lt11;j++)
{开关(映射[m].a[i][j])
{案例0:？printf("？);打破；
案例1:？printf("■")；打破；
？案例三:？printf("◎"；打破；
案例四:？printf("□")；打破；
案例五:？printf("♂"；打破；？//5是人
案例七:？printf("□")；打破；？//4+3盒子在目的地。
案例八:？printf("♂"；打破；//目的地5+3人
}
}
printf(" \ n \ t ")；
}
}
void gtxy(int x，int y)？//控制光标位置的函数
{ COORD coord
坐标。X = x
坐标。Y = y
SetConsoleCursorPosition(GetStdHandle(STD _ OUTPUT _ HANDLE)，coord)；
}
void start()？//开始游戏
{ int r，c；？//r，c用来记录人的下标。
for(int I = 0；我& lt9;i++)
{ for(int j = 0；j & lt11;j++)
？{if (map[m].a[I][j]= = 5 | |地图[m]。a[I][j]= = 8){ r = I；？c = j；}} //i j人的下标
}
char键；？
key = getch()；
开关(钥匙)
{大小写‘W’:
案例“w”:
案例72:
if(映射[m].a[r - 1][c] == 0||地图[m]。a [r - 1][c] == 3)
{ gtxy(2*c+8，r-1+3)；printf("♂"；？// gtxy(2*c+8，r-1+3)是输出到指定位置的字符。
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
地图[m]。a[r-1][c]+= 5；？地图[m]。a[r][c]-= 5；}
不然呢？if(映射[m].a [r - 1][c] == 4 ||地图[m]。a [r - 1][c] == 7)
{ if (map[m].a [r - 2][c] == 0 ||映射[m]。a[r-2][c]= 3)
{ gtxy(2*c+8，r-2+3)；printf("□")；gtxy(2*c+8，r-1+3)；printf("♂"；
if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
地图[m]。a[r-2][c]+= 4；？地图[m]。a[r-1][c]+= 1；
？地图[m]。a[r][c]-= 5；}
}破；
案例:
案例:
案例80:
if(映射[m].a [r + 1][c] == 0 ||地图[m]。a [r + 1][c] == 3)
？{ gtxy(2*c+8，r+1+3)；printf("♂"；
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
？if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
？地图[m]。a[r+1][c]+= 5；？地图[m]。a[r][c]-= 5；}
？else if (map[m].a [r + 1][c] == 4 ||地图[m]。a [r+ 1][c] == 7)
？{ if (map[m].a [r + 2][c] == 0 ||映射[m]。a [r + 2][c] == 3)
{ gtxy(2*c+8，r+2+3)；printf("□")；gtxy(2*c+8，r+1+3)；printf("♂"；
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
？if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
？地图[m]。a[r+2][c]+= 4；地图[m]。a[r+1][c]+= 1；
地图[m]。a[r][c]-= 5；}
？}破；
案例“A”:
案例“a”:
案例75:
if(映射[m].a [r ][c - 1] == 0 ||地图[m]。a [r ][c - 1] == 3)
？{ gtxy(2*(c-1)+8，r+3)；printf("♂"；
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
？if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
？地图[m]。a[r][c-1]+= 5；地图[m]。a[r][c]-= 5；}
？else if (map[m].a [r][c - 1] == 4 ||地图[m]。a [r][c - 1] == 7)
？{if (map[m].a [r ][c - 2] == 0 ||映射[m]。a [r ][c - 2] == 3)
{ gtxy(2*(c-2)+8，r+3)；printf("□")；gtxy(2*(c-1)+8，r+3)；printf("♂"；
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
？if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
？地图[m]。a[r][c-2]+= 4；地图[m]。a[r][c-1]+= 1；
？地图[m]。a[r][c]-= 5；}
？}破；
案例“D”:
案例“d”:
案例77:
if(映射[m].a [r][c + 1] == 0 ||地图[m]。a [r][c + 1] == 3)
？{ gtxy(2*(c+1)+8，r+3)；printf("♂"；
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
？if(映射[m].a[r ][c] == 8) {gtxy(2*c+8，r+3)；printf("◎"；}
？地图[m]。a[r][c+1]+= 5；？地图[m]。a[r][c]-= 5；}
？else if (map[m].a [r][c + 1] == 4 ||地图[m]。a [r][c + 1] == 7)
？{ if (map[m].a [r][c + 2] == 0 ||映射[m]。a [r][c + 2] == 3)
{ gtxy(2*(c+2)+8，r+3)；printf("□")；gtxy(2*(c+1)+8，r+3)；printf("♂"；
？if(映射[m].a[r ][c] == 5){gtxy(2*c+8，r+3)；printf("？);}
if(映射[m].a[r ][c] == 8){gtxy(2*c+8，r+3)；printf("◎"；}
地图[m]。a[r][c+2]+= 4；地图[m]。a[r][c+1]+= 1；
地图[m]。a[r][c]-= 5；}
？}破；
}
}
int ifwan()？//是否完成(1为0和否)
{ if(m==0){if(map[m].a[5][2]= = 7 & amp；& amp地图[m]。a[5][3]= = 7 & amp；& amp
？地图[m]。a[6][2]= = 7 & amp；& amp地图[m]。a[6][3]==7)返回1；}
if(m==1){if(map[m].a[5][2]= = 7 & amp；& amp地图[m]。a[6][2]= = 7 & amp；& amp
？地图[m]。a[7][2]==7)返回1；}
if(m==2){if(map[m].a[7][1]= = 7 & amp；& amp地图[m]。a[7][2]= = 7 & amp；& amp地图[m]。a[7][3]= = 7 & amp；& amp
地图[m]。a[7][4]= = 7 & amp；& amp地图[m]。a[7][5]==7)返回1；}
if(m==3){if(map[m].a[4][4]= = 7 & amp；& amp地图[m]。a[4][5]= = 7 & amp；& amp地图[m]。a[4][6]= = 7 & amp；& amp
地图[m]。a[5][4]= = 7 & amp；& amp地图[m]。a[5][5]= = 7 & amp；& amp地图[m]。a[5][6]==7)返回1；}
if(m==4){if(map[m].a[3][2]= = 7 & amp；& amp地图[m]。a[4][1]= = 7 & amp；& amp地图[m]。a[4][2]= = 7 & amp；& amp
地图[m]。a[5][1]= = 7 & amp；& amp地图[m]。a[5][2]==7)返回1；}
返回0；
}
int main()？//主函数
{ while (1)
{ system(" cls ")；
？dr map()；
？while (1)
？{ start()；
if(if wan()){ printf(" \ 007 ")；打破；}//完成后响铃。
？}
？m+= 1；
}
？返回0；
}