HTML5 canvas绘制的玫瑰花效果

有人用htML就写了一朵漂亮的玫瑰，是不是有点太骚气了。纯javascipt就弄出了玫瑰花，再次显示了HTML5的威力
警告：传说IE6内核是无法看到的，建议用Chrome或者Firefox。

玫瑰花效果：

实现代码：

复制代码代码如下:
!doctyPE html>
html>
head>
tITle> Love/title>
meta charset="utf-8" />
!--[if IE]>
script src="http://html5shiv.GOOGLEcode.COM/svn/trunk/html5.js"> /script>
![endif]-->
/head>
body>
canvas id="c"> /canvas>
script>
VAR b = document.body;
var c = document.getelementsbytagname('canvas')[0];
var a = c.getContext('2d');
document.body.clientWidth;
/script>
script>
// start of submission //
with (m = Math) C = cos, S = sin, P = pow, R = random; c.width = c.height = f = 500; h = -250; function p(a, b, c) { if (c > 60) return [S(a * 7) * (13 + 5 / (.2 + P(b * 4, 4))) - S(b) * 50, b * f + 50, 625 + C(a * 7) * (13 + 5 / (.2 + P(b * 4, 4))) + b * 400, a * 1 - b / 2, a]; A = a * 2 - 1; B = b * 2 - 1; if (A * A + B * B 1) { if (c > 37) { n = (j = c & 1) ? 6 : 4; o = .5 / (a + .01) + C(b * 125) * 3 - a * 300; w = b * h; return [o * C(n) + w * S(n) + j * 610 - 390, o * S(n) - w * C(n) + 550 - j * 350, 1180 + C(B + A) * 99 - j * 300, .4 - a * .1 + P(1 - B * B, -h * 6) * .15 - a * b * .4 + C(a + b) / 5 + P(C((o * (a + 1) + (B > 0 ? w : -w)) / 25), 30) * .1 * (1 - B * B), o / 1e3 + .7 - o * w * 3e-6] } if (c > 32) { c = c * 1.16 - .15; o = a * 45 - 20; w = b * b * h; z = o * S(c) + w * C(c) + 620; return [o * C(c) - w * S(c), 28 + C(B * .5) * 99 - b * b * b * 60 - z / 2 - h, z, (b * b * .3 + P((1 - (A * A)), 7) * .15 + .3) * b, b * .7] } o = A * (2 - b) * (80 - c * 2); w = 99 - C(A) * 120 - C(b) * (-h - c * 4.9) + C(P(1 - b, 7)) * 50 + c * 2; z = o * S(c) + w * C(c) + 700; return [o * C(c) - w * S(c), B * 99 - C(P(b, 7)) * 50 - c / 3 - z / 1.35 + 450, z, (1 - b / 1.2) * .9 + a * .1, P((1 - b), 20) / 4 + .05] } } setInterval('for(i=0; i1e4; i++)if(s=p(R(),R(),i%46/.74)){ z=s[2]; x=~~(s[0]*f/z-h); y=~~(s[1]*f/z-h); if(!m[q=y*f+x]|m[q]> z)m[q]=z,a.fillStyle="rgb("+~(s[3]*h)+","+~(s[4]*h)+","+~(s[3]*s[3]*-80)+")",a.fillRect(x,y,1,1)} ', 0)
// end of submission //
/script>
/body>
/html>

声明：本文内容由网友自发贡献，本站不承担相应法律责任。对本内容有异议或投诉，请联系2913721942#qq.com核实处理，我们将尽快回复您，谢谢合作！