C语言编程小游戏实在练习小项目
快来实操一下吧！（扣1送书本和例题资料）
直接上源码
#include<stdio.h>
#include<stdlib.h>
#include<getch.h>
int main(int argc,const char*argv[])
{
int cut=0;
int a[8][8]={
{0,0,3,3,3,3,0,0},
{0,0,3,5,5,3,0,0},
{0,0,3,0,5,3,0,0},
{0,3,0,0,4,5,3,0},
{3,3,0,4,0,0,3,3},
{3,0,0,3,4,4,0,3},
{3,0,0,2,0,0,0,3},
{3,3,3,3,3,3,3,3}};
int x=6,y=3;
for(;;)
{
int cunt=0;
system("clear");
for(int i=0;i<8;i )
{
for(int j=0;j<8;j )
{
if(a[i][j]==9)
{
cunt ;
}
switch(a[i][j])
{
case 0:printf(" ");break;
case 2:printf("@ ");break;//人
case 3:printf("# ");break;//墙
case 4:printf("$ ");break;//箱子
case 5:printf("o ");break;//目标位置
case 7:printf("@ ");break;
case 9:printf("$ ");break;
}
}
printf("
");
}
if(4==cunt)
{
printf("成功
步数：%d",cut);
return 0;
}
//根据数据的大小来判断当前的坐标上的情况
switch(getch())
{
case 183:
if(0!=x&&3==a[x-1][y])
{
a[x][y]=2;
}
else if(5==a[x-1][y]||0==a[x-1][y])
{
a[x][y]-=2;
a[x-1][y] =2;
x--;
cut ;
}
else if((5==a[x-2][y]||0==a[x-2][y])&&(4==a[x-1][y]||9==a[x-1][y]))
{
a[x-2][y] =4;
a[x-1][y]-=4;
a[x-1][y] =2;
a[x][y]-=2;
x--;cut ;
}break;
case 184:
if(0!=x&&3==a[x 1][y])
{
a[x][y]=2;
}
else if((5==a[x 1][y])||(0==a[x 1][y]))
{
a[x][y]-=2;
a[x 1][y] =2;
x ;cut ;
}
else if((5==a[x 2][y]||0==a[x 2][y])&&(4==a[x 1][y]||9==a[x 1][y]))
{
a[x 2][y] =4;
a[x 1][y]-=4;
a[x 1][y] =2;
a[x][y]-=2;
x ;cut ;
}break;
case 185:
if(0!=x&&3==a[x][y 1])
{
a[x][y]=2;
}
else if(5==a[x][y 1]||0==a[x][y 1])
{
a[x][y]-=2;
a[x][y 1] =2;
y ;cut ;
}
else if((0==a[x][y 2]||5==a[x][y 2])&&(4==a[x][y 1]||9==a[x][y 1]))
{
a[x][y 2] =4;
a[x][y 1]-=4;
a[x][y 1] =2;
a[x][y]-=2;
y ;cut ;
}break;
case 186:
if(0!=x&&3==a[x][y-1])
{
a[x][y]=2;
}
else if(5==a[x][y-1]||0==a[x][y-1])
{
a[x][y]-=2;
a[x][y-1] =2;
y--;cut ;
}
else if((0==a[x][y-2]||5==a[x-1][y])&&(4==a[x][y-1]||9==a[x][y-1]))
{
a[x][y-2] =4;
a[x][y-1]-=4;
a[x][y-1] =2;
a[x][y]-=2;
y--;cut ;
}break;
}
}
}
加油⛽⛽⛽！
