题目链接:http://codeforces.com/problemset/problem/93/B B. End of Exams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Students love to celebrate their holidays. Especially if the holiday is th
原因:js按照2进制来处理小数的加减乘除,在arg1的基础上 将arg2的精度进行扩展或逆扩展匹配,所以会出现如下情况. javascript(js)的小数点加减乘除问题,是一个js的bug如0.3*1 = 0.2999999999等,下面列出可以完美求出相应精度的四种js算法 function accDiv(arg1,arg2){ var t1=0,t2=0,r1,r2; try{t1=arg1.toString().split(".")[1].length}catch(e){} t
将精度高的浮点数转换成精度低的浮点数. 1.round()内置方法 这个是使用最多的,刚看了round()的使用解释,也不是很容易懂.round()不是简单的四舍五入的处理方式. For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus ndigits; if two multiples are equally close, roun