解决js计算精度问题

    // 除以
	 accDiv(arg1, arg2) {
			let t1 = 0,
				t2 = 0,
				r1, r2;
			try {
				t1 = arg1.toString().split(".")[1].length
			} catch (e) {}
			try {
				t2 = arg2.toString().split(".")[1].length
			} catch (e) {};
			r1 = Number(arg1.toString().replace(".", ""))
			r2 = Number(arg2.toString().replace(".", ""))
			return (r1 / r2) * Math.pow(10, t2 - t1);
		}，
		// 乘以
		accMul(arg1, arg2) {

			let m = 0,
				s1 = arg1.toString(),
				s2 = arg2.toString();
			try {
				m += s1.split(".")[1].length
			} catch (e) {}
			try {
				m += s2.split(".")[1].length
			} catch (e) {}
			return Number(s1.replace(".", "")) * Number(s2.replace(".", "")) / Math.pow(10, m)
		},
		// 减
		accSub(arg1, arg2) {
			let r1, r2, m, n;
			try {
				r1 = arg1.toString().split(".")[1].length
			} catch (e) {
				r1 = 0
			}
			try {
				r2 = arg2.toString().split(".")[1].length
			} catch (e) {
				r2 = 0
			}
			m = Math.pow(10, Math.max(r1, r2));
			//动态控制精度长度 
			n = (r1 >= r2) ? r1 : r2;
			return ((arg1 * m - arg2 * m) / m).toFixed(n);
		},
        //加法 
		accAdd(arg1, arg2) {
			let r1, r2, m;
			try {
				r1 = arg1.toString().split(".")[1].length
			} catch (e) {
				r1 = 0
			}
			try {
				r2 = arg2.toString().split(".")[1].length
			} catch (e) {
				r2 = 0
			}
			m = Math.pow(10, Math.max(r1, r2))
			return (arg1 * m + arg2 * m) / m
		}