使用js写点击一个事件使页面返回顶部以及控制一个元素在右下角的固定位置的方法

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>无标题文档</title>
<style type="text/css">
body{margin:0; padding:0}
#to_top{
30px;
height:40px;
padding:20px;
font:14px/20px arial;
text - align: center;
background: #06c;
position: absolute;
cursor: pointer;
color: #fff;
}
</style>
</head>

<body style="height:1000px;">
             <h1>返回顶部</h1 >
    <div id = "to_top"> 返回顶部
          </div>
</body>
</html>
<script type="text/javascript">
window.onload = function(){
var oTop = document.getElementById("to_top");
var screenw = document.documentElement.clientWidth || document.body.clientWidth;
var screenh = document.documentElement.clientHeight || document.body.clientHeight;
oTop.style.left = screenw - oTop.offsetWidth +"px";
oTop.style.top = screenh - oTop.offsetHeight + "px";
window.onscroll = function(){
   var scrolltop = document.documentElement.scrollTop || document.body.scrollTop;
   oTop.style.top = screenh - oTop.offsetHeight + scrolltop +"px";
  }
oTop.onclick = function(){
  document.documentElement.scrollTop = document.body.scrollTop =0;
}
}
/**jquery的写法实现返回顶部
jQuery("#to_top").click(function(){
  jQuery("html,body").animate({'scrollTop':0},100)
})
**/

</script>