基于C语言的小游戏合集

小游戏之推箱子

#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("\n");
	   }
	   if(4==cunt)
	   {
		   printf("成功\n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;
	  }
}		
}

小游戏之走迷宫

#include<stdio.h>
#include<getch.h>
#include<stdlib.h>
#include<time.h>
int main(int argc,const char*argv[])
{
	//构造迷宫地图
	char maze[10][10]={
	                   {'#','#','#','#','#','#','#','#','#','#'},
	                   {'#','@','#','#','#','#','#','#','#','#'},
	                   {'#',' ','#','#','#','#','#','#','#','#'},
	                   {'#',' ','#','#','#','#','#','#','#','#'},
	                   {'#',' ','#','#','#',' ',' ',' ',' ','#'},
	                   {'#',' ',' ',' ','#',' ','#','#',' ',' '},
	                   {'#','#','#',' ',' ',' ','#','#','#','#'},
	                   {'#','#','#','#','#','#','#','#','#','#'},
	                   {'#','#','#','#','#','#','#','#','#','#'},
	                   {'#','#','#','#','#','#','#','#','#','#'}};	
	   time_t t1=time(NULL);
	   int x=1,y=1;			 
		for(;;)
		{
	     	system("clear");
		    for(int i=0;i<10;i++)
		    {
			    for(int j=0;j<10;j++)
		 	    {
			        printf("%c ",maze[i][j]);
		        }
				printf("\n");
		    }
			if(x==5&&y==9)
			{
		    	printf("通过\n");
				printf("通过时间%u",time(NULL)-t1);
			return 0;
			}
			//获取键值
			switch(getch())
			{
				case 183:
						 if(x>0&& ' '==maze[x-1][y])
					     {
						 maze[x][y]=' ';
						 maze[x-1][y]='@';
						 x--;
					     }
					 break;
				case 184:
				     if(x<9&& ' '==maze[x+1][y])
					 {
						 maze[x][y]=' ';
						 maze[x+1][y]='@';
						 x++;
					 }
					 break;
				case 185:
				     if(y<9&& ' '==maze[x][y+1])
					 {
						 maze[x][y]=' ';
						 maze[x][y+1]='@';
						 y++;
					 }
					 break;
				case 186:
				     if(y>0&&' ' ==maze[x][y-1])
					 {
						 maze[x][y]=' ';
						 maze[x][y-1]='@';
						 y--;
					 }
					 break;
			}
	}
}

小游戏之五子棋

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <getch.h>
// 棋盘
char board[15][15];
// 棋子坐标
char kx = 7 , ky = 7;
// 角色
char role = '@';
// 显示棋盘
void show_board(void)
{
	system("clear");
	for(int i=0; i<15; i++)
	{
		for(int j=0; j<15; j++)
		{
			if(board[i][j])
			{
				printf(" %c",board[i][j]);
			}
			else
			{
				printf(" *");
			}
		}
		printf("\n");
	}
}
// 落子
void get_key(void)
{
	printf(" -----------请%c落子-----------\n",role);
	for(;;)
	{
		//通过光标下棋
		printf("\33[%hhd;%hhdH",kx+1,(ky+1)*2);
		switch(getch())
		{
			case 183: kx>0 && kx--; break;
			case 184: kx<14 && kx++; break;
			case 185: ky<14 && ky++; break;
			case 186: ky>0 && ky--; break;
			case 10: if(!board[kx][ky])
			{
				board[kx][ky] = role;
				return;
			}
		}
	}
}
int count_eqkey(int ox,int oy)
{
	int count = 0;
	for(int x=kx+ox,y=ky+oy; x>=0 && x<15 && y>=0 && y<15 &&
		board[x][y]==board[kx][ky]; x+=ox,y+=oy)
	{
		count++;
	}
	return count;
}
// 检查五子
bool check_board(void)
{
	if(count_eqkey(0,-1)+count_eqkey(0,1) >= 4)
		return true;
	if(count_eqkey(-1,0)+count_eqkey(1,0) >= 4)
		return true;
	if(count_eqkey(-1,-1)+count_eqkey(1,1) >= 4)
		return true;
	if(count_eqkey(-1,1)+count_eqkey(1,-1) >= 4)
		return true;
	return false;
}
int main()
{
	for(;;)
	{
		show_board();
		get_key();		
		if(check_board())
		{
			show_board();
			printf("恭喜%c胜利,游戏结束!\n",role);
			return 0;
		}
		role = '@'==role?'#':'@';
	}
}