[转]逆天的俄罗斯方块，只需要60行代码

知乎上看到这个逆天的代码，感觉我不会写js. 

https://www.zhihu.com/question/44213758/answer/97448141?group_id=708529590292713472

<!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>

posted @ 2016-04-27 21:55  CooMark  阅读(1852)  评论(0)  编辑  收藏  举报