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60行js代碼實現俄羅斯方塊

 這是我之前網上看到的,很牛逼的一位大神寫的,一直膜拜中

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 <!doctype html><html><head></head><body> <div id="box" style="width:252px;font:25px/25px 宋體;background:#000;color:#9f9;border:#999 20px ridge;text-shadow:2px 3px 1px #0f0;"></div> <script> var map=eval("["+Array(23).join("0x801,")+"0xfff]"); var tatris=[[0x6600],[0x2222,0xf00],[0xc600,0x2640],[0x6c00,0x4620],[0x4460,0x2e0,0x6220,0x740],[0x2260,0xe20,0x6440,0x4700],[0x2620,0x720,0x2320,0x2700]]; var keycom={"38":"rotate(1)","40":"down()","37":"move(2,1)","39":"move(0.5,-1)"}; var dia, pos, bak, run; function start(){ dia=tatris[~~(Math.random()*7)]; bak=pos={fk:[],y:0,x:4,s:~~(Math.random()*4)}; rotate(0); } function over(){ document.onkeydown=null; clearInterval(run); alert("GAME OVER"); } function update(t){ bak={fk:pos.fk.slice(0),y:pos.y,x:pos.x,s:pos.s}; if(t) return; for(var i=0,a2=""; i<22; i++) a2+=map[i].toString(2).slice(1,-1)+"<br/>"; for(var i=0,n; i<4; i++) if(/([^0]+)/.test(bak.fk[i].toString(2).replace(/1/g,"u25a1"))) a2=a2.substr(0,n=(bak.y+i+1)*15-RegExp.$_.length-4)+RegExp.$1+a2.slice(n+RegExp.$1.length); document.getElementById("box").innerHTML=a2.replace(/1/g,"u25a0").replace(/0/g,"u3000"); } function is(){ for(var i=0; i<4; i++) if((pos.fk[i]&map[pos.y+i])!=0) return pos=bak; } function rotate(r){ var f=dia[pos.s=(pos.s+r)%dia.length]; for(var i=0; i<4; i++) pos.fk[i]=(f>>(12-i*4)&15)<<pos.x; update(is()); } function down(){ ++pos.y; if(is()){ for(var i=0; i<4 && pos.y+i<22; i++) if((map[pos.y+i]|=pos.fk[i])==0xfff) map.splice(pos.y+i,1), map.unshift(0x801); if(map[1]!=0x801) return over(); start(); } update(); } function move(t,k){ pos.x+=k; for(var i=0; i<4; i++) pos.fk[i]*=t; update(is()); } document.onkeydown=function(e){ eval(keycom[(e?e:event).keyCode]); }; start(); run=setInterval("down()",400); </script></body></html>
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