俄罗斯方块

来源：互联网发布：淘宝口令在哪里找到编辑：程序博客网时间：2024/06/06 10:40

今天在知乎上看见一个60行敲出俄罗斯方块的代码，分享.

<!DOCTYPE html><html><head>    <meta http-equiv="Content-Type" content="text/html; charset=GBK"></head><body>    <div id="box" style="width: 252px; font: 25px/25px 宋体; background: #000; color: #9f9; 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>

粘贴复制为html文件就可以运行
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地址：

https://www.zhihu.com/question/44213758/answer/97448141

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