var div=document.getElementById("father");
var children=div.document.getElementsByTagName("div");
for(var i=0;i<children.length;i++){
    children[i].style.width=parseInt(children[i].style.width)+i*10;
    children[i].style.height=parseInt(children[i].style.height)+i*10;
}这样么?

解决方案 »

  1.   

    <html>
    <head>
    <title></title>
    <style>
    #Father div {
        float: left;
        margin: 5px;
        border: solid 1px #000;
        width: 20px;
        height: 20px;
    }
    </style>
    <script language="JavaScript">
    function ZoomIn()
    {
        var father = document.getElementById("Father");
        for(var i = 0; i < father.children.length; i++)
        {
            father.children[i].style.width = 20 + i * 5;
            father.children[i].style.height = 20 + i * 5;
        }
    }
    </script>
    </head>
    <body>
    <div id="Father">
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    <div></div>
    </div>
    <p style="clear:both;"><button onclick="ZoomIn()">放大</button></p>
    </body>
    </html>
      

  2.   

    任何递增函数都可以
    设x∈[n, m] n,m的值根据定义域和实际变化情况定
    采样距离△x =(m-n)/10
    比如
    线性函数 y = ax + b (a>0) 等量增长
    指数函数 y = x^a 越长越快
    对数函数 y = ln(x) 越长越慢
      

  3.   

    超级大笨狼,灌水专用脚本:
     <SCRIPT LANGUAGE="JavaScript">
      Draw_Triangle(40);
      Draw_Line(40);
    function Draw_Triangle(line)
    {
    var chars="";
    for(i=1;i<=line;i++)
    {
    var blank="";
    var star="";
    for(j=0;j<i;j++)
    {
    star +="*&nbsp;" ;

    for(k=0;k<(line-i);k++)
    {
    blank +="&nbsp";  

    chars+=blank + star + "<br/>";

    }
    document.write(chars);
    }function Draw_Line(line)
    {
    var chars="";
    for(i=1;i<=line;i++)
    {
    var blank="";
    var star="";
    for(j=0;j<line/2;j++)
    {
    star +="*&nbsp;" ;

    for(k=0;k<line/2;k++)
    {
    blank +="&nbsp";  

    chars+=blank + star + "<br/>";

    }
    document.write(chars);
    }
     </SCRIPT>
                                           * 
                                          * * 
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