欧美大屁股bbbbxxxx,狼人大香伊蕉国产www亚洲,男ji大巴进入女人的视频小说,男人把ji大巴放进女人免费视频,免费情侣作爱视频

歡迎來到入門教程網(wǎng)!

C語言

當(dāng)前位置:主頁 > 軟件編程 > C語言 >

VC++ 6.0 C語言實(shí)現(xiàn)俄羅斯方塊詳細(xì)教程

來源:本站原創(chuàng)|時(shí)間:2020-01-10|欄目:C語言|點(diǎn)擊: 次

今天把我之前寫的大作業(yè)分享一下吧,并教你們?nèi)绾螌?shí)現(xiàn),希望你們看了前面的教程也能自己寫一個(gè)。

1.要先下載一個(gè) graphics.h 的頭文件來繪圖。

2.初始化窗口:initgraph(x, y);這是先創(chuàng)建一個(gè)窗口的函數(shù),以左上角為(0,0),向右為x軸,向下為y軸,其中x表示長x個(gè)單位,y表示寬y個(gè)單位。

3.關(guān)閉圖像窗口:closegraph();結(jié)束時(shí)用來關(guān)閉用的。

4.按任意鍵繼續(xù):getch();這個(gè)就和getchar();差不多,為了防止以運(yùn)行完就關(guān)了,這樣能停頓一下,他的頭文件是:conio.h 。

5.畫線:line(x1, y1, x2, y2);在你創(chuàng)建的那么窗口里以(x1,y1)和(x2,y2)為兩個(gè)端點(diǎn)畫線。

6.畫矩形:rectangle(x1,y1,x2,y2);以(x1,y1)和(x2,y2)為對(duì)角畫一個(gè)矩形。

7.畫圓:circle(x,y,r);以(x,y)為圓點(diǎn),r為半徑畫圓。

8.顏色:setcolor(x);用來設(shè)置顏色的,其中x是你要設(shè)置的顏色,可以填這16種:黑 BLACK、藍(lán) BLUE、綠 GREEN、青 CYAN、紅 RED、紫 MAGENTA、棕 BROWN、淺灰 LIGHTGRAY、深灰 DARKGRAY、亮藍(lán) LIGHTBLUE、亮綠 LIGHTGREEN、亮青 LIGHTCYAN、亮紅 LIGHTRED、亮紫 LIGHTMAGENTA、黃 YELLOW、白 WHITE;當(dāng)然,你也可以根據(jù)光的三原色來調(diào)自己喜歡的顏色,方法是:setcolor(RGB(x,y,z));其中RGB分別代表紅綠藍(lán),對(duì)應(yīng)的x,y,z是你選的該顏色的多少,范圍是[0,255]。

根據(jù)上面的這些就能畫出很多圖形了.比如:

# include <graphics.h> 
# include <conio.h> 
 void main() 
 {   
   initgraph(640,480); 
   setcolor(BROWN); 
   line(200,200,300,130); 
   line(400,200,300,130); 
   rectangle(200,200,400,380); 
   setcolor(RGB(100,150,200)); 
   circle(255,260,20); 
   circle(345,260,20); 
   rectangle(280,300,310,360); 
   getch(); 
   closegraph(); 
 } 

還有這個(gè),畫一個(gè)圓,圓心是(320,240),半徑r=200,根據(jù)角度的公式畫一個(gè)圓:

# include <graphics.h> 
# include <conio.h> 
# include <math.h> 
# define PI 3.1415926 
 void main() 
 { 
   initgraph(640, 480); 
   int x,y,r=200,c; 
   double a; 
   for(a=0; a<PI*2; a+=0.0001) 
   { 
     x=(int)(r*cos(a)+320+0.5); 
     y=(int)(r*sin(a)+240+0.5); 
     c=(int)(a*255/(2*PI)); 
     setcolor(RGB(c,200-c/2,100+c/2)); 
     line(320,240,x,y); 
   } 
   getch(); 
   closegraph(); 
 } 

9.停頓:Sleep(x);停頓x/1000秒。

有了Sleep();函數(shù),就能讓線動(dòng)起來了,原理就是:先畫一條線,然后再畫一條黑色的線覆蓋原來的那條,然后再畫一條線,這樣不斷的畫線能行了。

# include <graphics.h> 
# include <conio.h> 
 void main() 
 { 
   int i,y; 
   initgraph(640,480); 
   for(y=0;y<480-2;y++) 
   { 
     setcolor(RGB(y,125,y/2%256)); 
     line(0,y,639,y); 
     line(0,y+2,639,y+2); 
     Sleep(10); 
     setcolor(BLACK); 
     line(0,y,639,y); 
   } 
   getch(); 
   closegraph(); 
 } 

10.隨機(jī)數(shù)發(fā)生器:srand();使用方法:srand(time(NULL));使用時(shí)要有頭文件time.h。

11.隨機(jī)函數(shù):rand();隨機(jī)生成一個(gè)數(shù),頭文件是:stdlib.h,比如:

# include <stdio.h> 
# include <time.h> 
# include <graphics.h> 
 void main() 
 { 
   int t=10; 
   while(t--) 
   { 
     srand(time(NULL)); 
     printf("%d\n",rand()); 
     Sleep(1000); 
   } 
} 

12.判斷鍵盤是否有輸入:kbhit();如果有的話返回1,否則返回0.

13.方向鍵:方健健的ASCII值我們不知道,普通的getchar();也不能輸入,但是getch();通過運(yùn)行程序可以發(fā)現(xiàn)方向比較特殊,分別是:上 224+72、下 224+80、左 224+75、右 224+77,就是說他們是由兩個(gè)字符組成的,所以判斷上下左右時(shí)就先判斷if(kbhit()),然后判斷if(getch()==224),如果是的話在判斷if(getch()==72),是的話就是上,下左右同理。

# include <stdio.h> 
# include <graphics.h> 
# include <conio.h> 
 void main() 
 { 
   int key; 
   while(1) 
   { 
     if(kbhit()) 
     { 
       key=getch(); 
       if(key==224) 
         key=getch(); 
       printf("%d",key); 
     } 
     else 
       printf("*"); 
     Sleep(500); 
   } 
 } 

14.輸出:outtextxy(x,y,s);在坐標(biāo)為(x,y)的地方輸出字符串s。

15.數(shù)字轉(zhuǎn)字符串:sprintf(s,"%d",a);將數(shù)字a,轉(zhuǎn)化為字符串s,頭文件是stdio.h。

16.設(shè)置字體:setfont(x,y,"字體");其中x表示你你要輸出的字的大小,y會(huì)改變形狀一般都是設(shè)為0,"字體"為你想要輸出的字體,比如:setfont(40,0,"楷體");

# include <graphics.h> 
# include <conio.h> 
# include <stdio.h> 
 void main() 
 { 
   char s[10]; 
   initgraph(640,480); 
   setfont(42,0,"隸體"); 
   outtextxy(200,200,"俄羅斯方塊"); 
   sprintf(s,"%d",2015); 
   outtextxy(250,250,s); 
   getch(); 
   closegraph(); 
 } 

上面這些都學(xué)會(huì)了的話,差不多就能夠自己完成編寫一個(gè)俄羅斯方塊了,下面分享一下我寫的代碼。

# include <stdio.h> 
# include <time.h> 
# include <conio.h> 
# include <graphics.h> 
 int a,visit[30][15],Color[30][15],minX=30,minY=30,maxX=310,maxY=610,speed=500,score=0,rank=0; 
 int color[]={GREEN,CYAN,MAGENTA,BROWN,LIGHTGRAY,LIGHTBLUE,LIGHTGREEN,LIGHTRED,YELLOW}; 
 int block[36][5][5]= 
 { 
   {0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,1,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0}, 
   {0,0,1,1,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,1,1,0}, 
   {0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,1,0,1,0,0,0,0,0}, 
   {0,1,1,0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,0,0,0,1,1,0,0}, 
   {0,0,0,0,0,1,0,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,0,1,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,1,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,1,1,0,0,0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, 
   {0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0} 
 }; 
 void Interface();//界面的設(shè)計(jì) 
 void nextblock();//生成下一個(gè)方塊 
 void newblock();//將下一個(gè)放到主方框中 
 void move(int x,int y,int A);//移動(dòng)方塊 
 int couldmove(int x0,int y0,int A,char dir,int next);//判斷能否移動(dòng) 
 int couldrotate(int x0,int y0,int A,int next);//判斷能否旋轉(zhuǎn) 
 void mark(int x0,int y0,int A,int next);//標(biāo)記方塊固定后的位置 
 void full();//判斷有沒有能夠消行的 
 void down(int x);//有消行的將上面的方塊下移一行 
 int gameover();//判斷游戲是否結(jié)束 
 int main() 
 { 
   int i,j; 
   Interface(); 
   nextblock(); 
   Sleep(500); 
   memset(visit,0,sizeof(visit)); 
   while(1) 
   { 
   newblock(); 
   full(); 
   if(gameover()) 
     break; 
   } 
   getch(); 
   closegraph(); 
   return 0; 
 } 
 void Interface() 
 { 
   int i,j; 
   char str[10]; 
   initgraph(550, 660); 
   HWND hwnd = GetHWnd(); 
   SetWindowText(hwnd, "NEW俄羅斯方塊         制作:丶阿明"); 
   Sleep(2000); 
   setfont(40,0,"隸體"); 
   setcolor(RED); 
   outtextxy(145,200,"NEW"); 
   setfont(40,0,"楷體"); 
   for(i=0;i<=255;i++) 
     for(j=0;j<=255;j++) 
     { 
       setcolor(RGB((i+j+128)%255,i,j)); 
       outtextxy(205,200,"俄羅斯方塊!"); 
     } 
     Sleep(2000); 
     setcolor(YELLOW); 
     setfont(22,0,"隸體"); 
     outtextxy(310,300,"←_←"); 
     setfont(22,0,"楷體"); 
     outtextxy(180,300,"這逼裝的可以"); 
     setcolor(LIGHTMAGENTA); 
     Sleep(2500); 
     outtextxy(175,300,"請(qǐng)按任意鍵開始游戲!"); 
     getch(); 
     cleardevice(); 
     setcolor(WHITE); 
     rectangle(29,29,334,633); 
     rectangle(28,28,335,634); 
     rectangle(370,50,515,195); 
     setfont(24,0, "楷體"); 
     setcolor(LIGHTGRAY); 
     outtextxy(405,215,"下一個(gè):"); 
     setcolor(RED); 
     outtextxy(405,280,"分?jǐn)?shù):"); 
     sprintf(str,"%d",score); 
     outtextxy(415,310,str); 
     outtextxy(405,375,"等級(jí):"); 
     sprintf(str,"%d",rank); 
     outtextxy(425,405,str); 
     setfont(22,0, "楷體"); 
     setcolor(LIGHTBLUE); 
     outtextxy(390,475,"操作說明:"); 
     outtextxy(390,500,"↑: 旋轉(zhuǎn)"); 
     outtextxy(390,525,"↓: 下降"); 
     outtextxy(390,550,"←: 左移"); 
     outtextxy(390,575,"→: 右移"); 
     outtextxy(390,600,"空格: 暫停"); 
 } 
 void nextblock() 
 { 
   int i,j,x=391,y=71; 
   setfont(23,0,"楷體"); 
   setcolor(BLACK); 
   for(i=0;i<5;i++) 
     for(j=0;j<5;j++) 
       outtextxy(x+20*j,y+20*i,"■"); 
     srand(time(NULL)); 
     a=rand()%9; 
     setcolor(color[a]); 
     for(i=0;i<5;i++) 
       for(j=0;j<5;j++) 
         if(block[4*a][i][j]==1) 
           outtextxy(x+20*j,y+20*i,"■"); 
 } 
 void newblock() 
 { 
   int i,j,x=130,y=30,A=a; 
   if(A==0||A==1||A==6) 
     y-=60; 
   else if(A==2||A==3||A==5||A==7||A==8) 
     y-=40; 
   else 
     y-=80; 
   setfont(23,0,"楷體"); 
   setcolor(color[A]); 
   for(i=0;i<5;i++) 
     for(j=0;j<5;j++) 
       if(block[4*A][i][j]==1&&y+20*i>=30) 
         outtextxy(x+20*j,y+20*i,"■"); 
       Sleep(100); 
       nextblock(); 
       move(x,y,A); 
 } 
 void move(int x,int y,int A) 
 { 
   int i,j,f,k=0,next=0,Speed=speed,key,nex; 
   while(1) 
   { 
     f=couldmove(x,y+k,A,'x',next); 
     if(f==0) 
     { 
       mark(x,y+k,A,next); 
       break; 
     } 
     if(kbhit()) 
     { 
       key=getch(); 
       if(key==32) 
         getch(); 
     } 
     setcolor(BLACK); 
     for(i=0;i<5;i++) 
       for(j=0;j<5;j++) 
         if(block[4*A+next][i][j]==1&&y+20*i+k>=30) 
           outtextxy(x+20*j,y+20*i+k,"■"); 
         if(kbhit()) 
         { 
           key=getch(); 
           if(key==72) 
           { 
             nex=(next+1)%4; 
             if(couldrotate(x,y+k,A,nex)) 
               next=nex; 
           } 
           else if(key==80) 
             Speed=50; 
           else if(key==75&&couldmove(x,y+k+20,A,'z',next)) 
             x-=20; 
           else if(key==77&&couldmove(x,y+k+20,A,'y',next)) 
             x+=20; 
         } 
         setcolor(color[A]); 
         for(i=0;i<5;i++) 
           for(j=0;j<5;j++) 
             if(block[4*A+next][i][j]==1&&y+20*i+k+20>=30) 
               outtextxy(x+20*j,y+20*i+k+20,"■"); 
             Sleep(Speed); 
             k+=20; 
       } 
 } 
 int couldmove(int x0,int y0,int A,char dir,int next) 
 { 
   int i,j,x,y,f=1; 
   x=(y0-minY)/20; 
   y=(x0-minX)/20; 
   if(dir=='x') 
   { 
     for(i=0;i<5;i++) 
       for(j=0;j<5;j++) 
       if(block[4*A+next][i][j]==1&&(x+i+1==30||(x+i+1>=0&&x+i+1<30&&y+j>=0&&y+j<15&&visit[x+i+1][y+j]==1))) 
         f=0; 
   } 
   else if(dir=='z') 
   { 
     for(i=0;i<5;i++) 
       for(j=0;j<5;j++) 
         if(block[4*A+next][i][j]==1&&(y+j==0||(x+i>=0&&x+i<30&&y+j-1>=0&&y+j-1<15&&visit[x+i][y+j-1]==1))) 
           f=0; 
   } 
   else if(dir=='y') 
   { 
     for(i=0;i<5;i++) 
       for(j=0;j<5;j++) 
         if(block[4*A+next][i][j]==1&&(y+j+1==15||(x+i>=0&&x+i<30&&y+j+1>=0&&y+j+1<15&&visit[x+i][y+j+1]==1))) 
           f=0; 
   } 
   return f; 
 } 
 int couldrotate(int x0,int y0,int A,int next) 
 { 
   int i,j,x,y,f=1; 
   x=(y0-minY)/20; 
   y=(x0-minX)/20; 
   if(!couldmove(x0,y0,A,'x',next)) 
     f=0; 
   for(i=0;i<5;i++) 
     for(j=0;j<5;j++) 
       if(block[4*A+next][i][j]==1&&(y+j==-1||y+j==15||(x+i>=0&&x+i<30&&y+j>=0&&y+j<15&&visit[x+i][y+j]==1))) 
         f=0; 
       return f; 
 } 
 void mark(int x0,int y0,int A,int next) 
 { 
   int i,j,x,y; 
   x=(y0-minY)/20; 
   y=(x0-minX)/20; 
   for(i=0;i<5;i++) 
     for(j=0;j<5;j++) 
       if(block[4*A+next][i][j]==1) 
       { 
         visit[x+i][y+j]=1; 
         Color[x+i][y+j]=color[A]; 
       } 
 } 
 void full() 
 { 
   int i,j,k=0; 
   char str[10]; 
   for(i=29;i>=0;i--) 
   { 
     for(j=0;j<15;j++) 
       if(visit[i][j]==0) 
         break; 
       if(j==15) 
       { 
         k++; 
         down(i); 
         i++; 
       } 
   } 
   setcolor(RED); 
   score+=10*k; 
   sprintf(str,"%d",score); 
   outtextxy(415,310,str); 
   rank=(score/50); 
   sprintf(str,"%d",rank); 
   outtextxy(425,405,str); 
   speed=500-100*rank; 
   if(speed==0) 
     speed=100; 
   if(score>=300) 
     speed=75; 
 } 
 void down(int x) 
 { 
   int i,j; 
   for(i=x;i>0;i--) 
     for(j=0;j<15;j++) 
       if(visit[i-1][j]==1) 
       { 
         visit[i][j]=visit[i-1][j]; 
         Color[i][j]=Color[i-1][j]; 
         setcolor(Color[i-1][j]); 
         outtextxy(minX+20*j,minY+20*i,"■"); 
       } 
       else 
       { 
         visit[i][j]=visit[i-1][j]; 
         setcolor(BLACK); 
         outtextxy(minX+20*j,minY+20*i,"■"); 
       } 
       setcolor(BLACK); 
       for(j=0;j<15;j++) 
       { 
         visit[0][j]=0; 
         outtextxy(minX+20*j,minY,"■"); 
       } 
 } 
 int gameover() 
 { 
   int i,j; 
   for(i=0;i<15;i++) 
     if(visit[0][i]==1) 
     { 
       for(i=0;i<=7;i++) 
       { 
         setcolor(RGB(35*i,255-35*i,255)); 
         for(j=i;j<30-i;j++,Sleep(20)) 
           outtextxy(30+20*i,30+20*j,"■"); 
         for(j=i;j<15-i;j++,Sleep(20)) 
           outtextxy(30+20*j,30+20*(29-i),"■"); 
         for(j=30-1-i;j>=i;j--,Sleep(20)) 
           outtextxy(30+20*(14-i),30+20*j,"■"); 
         for(j=15-1-i;j>=i;j--,Sleep(20)) 
           outtextxy(30+20*j,30+20*i,"■"); 
       } 
       setcolor(WHITE); 
       setfont(45,0,"隸體"); 
       outtextxy(75,300,"GAME OVER!"); 
       Sleep(5000); 
       return 1; 
     } 
     return 0; 
 } 

我設(shè)置的一行能放15個(gè)小格子,一共能放30行,二維數(shù)組visit[30][15],來判斷某個(gè)位置是否有方塊,比如visit[29][5]=1表示第30行第6個(gè)方塊的位置有方塊了,0表示沒方塊,Color[30][15]用來記錄某個(gè)位置的方塊的顏色。minX=30,minY=30表示左上角的方塊也就是第1行第1列的方塊對(duì)應(yīng)在界面的坐標(biāo),maxX,maxY同理。speed表示方塊的下落速度,一開始初始化為500,沒0.5s往下落一行,score記錄分?jǐn)?shù),每消一行加10分,rank表示等級(jí),每50分等級(jí)+1,且速度減100,當(dāng)速度減到100時(shí),就不變了,只有當(dāng)分?jǐn)?shù)超過300速度才會(huì)減為75,且以后都不會(huì)再減少了。

我一個(gè)設(shè)計(jì)了9種方塊,每種方塊都有4種形態(tài),每順時(shí)針旋轉(zhuǎn)90°就有一種形態(tài),所以一共是36個(gè)方塊。每個(gè)方塊都用5*5的二維數(shù)組來表示,36種就是36個(gè)二維數(shù)組,所以用block[36][5][5]來表示,block[k][5][5],(k<36&&k%4==0)表示每種方塊的初始形態(tài),k+1,+2,+3,分別表示第k種方塊選擇90°、180°、270°之后的形態(tài)。比如block[1][5][5]保存的是第1種方塊旋轉(zhuǎn)90°的方塊,其中block[1][i][j]=1表示這個(gè)方塊第i+1行,j+1列有一個(gè)小方塊,=0代表沒有,所以我的block[36][5][5],有36行,一行25個(gè)數(shù)列,表示一種方塊。一維數(shù)組color[]依次表示9種方塊的顏色。

附上效果圖:

以上就是本文的全部內(nèi)容,希望對(duì)大家的學(xué)習(xí)有所幫助,也希望大家多多支持我們。

上一篇:c++創(chuàng)建二維動(dòng)態(tài)數(shù)組與內(nèi)存釋放問題

欄    目:C語言

下一篇:C語言基于貪心算法解決裝箱問題的方法

本文標(biāo)題:VC++ 6.0 C語言實(shí)現(xiàn)俄羅斯方塊詳細(xì)教程

本文地址:http://mengdiqiu.com.cn/a1/Cyuyan/762.html

網(wǎng)頁制作CMS教程網(wǎng)絡(luò)編程軟件編程腳本語言數(shù)據(jù)庫服務(wù)器

如果侵犯了您的權(quán)利,請(qǐng)與我們聯(lián)系,我們將在24小時(shí)內(nèi)進(jìn)行處理、任何非本站因素導(dǎo)致的法律后果,本站均不負(fù)任何責(zé)任。

聯(lián)系QQ:835971066 | 郵箱:835971066#qq.com(#換成@)

Copyright © 2002-2020 腳本教程網(wǎng) 版權(quán)所有