c# 算法

[dongyangmoney]

现在有两个数例如：
double a = 1.1;
double b = 100.11;
要求实现 根据a的小数位数去拆分b
返回两个结果
一个为跟 a 有相同小数位数的数，一个为原数减去前面获得的那个有相同小数位数的数，
如上例中应该返回 100.1 和 0.01

(注意a 或b 无小数的情况)

[bloodish]

            double a = 1.1;
            double b = 100.11;
       
            double ret1 = (long)b + (a-(long)a);
            double ret2 = b - ret1;

[龍月]

情况比较多， 把所有情况列出来 判断吧 此消息通过 【CSDN论坛 Winform测试版】 回复！ 龙月.NET的博客

[边城的刀声]

double a = 1.1;
            double b = 100.11;

            String strA = a.ToString();
            Int32 dotNumLenth = strA.IndexOf('.') == -1 ? 0 : strA.Split('.')[1].Length;

            double result1 = Convert.ToDouble(b.ToString(String.Format(@"N{0}", dotNumLenth)));
            double result2 = b - result1;
            
            Console.WriteLine(result1);
            Console.WriteLine(result2);

[HolyPlace]

            double a = 1.01;
            double b = 100.11;

            string strA = a.ToString();
            string strB = string.Empty;
            string strC = string.Empty;
            strB = b.ToString().Substring(0, b.ToString().IndexOf('.') + (strA.Length - strA.IndexOf('.')));

            strC = "0.0";
            for (int i = 1; i < b.ToString().Length - b.ToString().IndexOf('.') -1; i++)
            {
                strC += "0";
            }
            strC = (b - double.Parse(strB)).ToString(strC);

strB 和 strC就是

[redmudwashcompanion]

public void Compute()
{
double dblData1 = 1.1;
double dblData2 = 100.11;
double dblData3 = 0;
double dblData4 = 0;
string strValue = "";
int intLength = 0;

int intIndex = 0;
int intIndex1 = 0;

intIndex = dblData1.ToString().IndexOf('.');
if (intIndex == -1)
intLength = 0;
else
{
intIndex++;

if (intIndex >= 0)
intLength = dblData1.ToString().Length - intIndex;
else
intLength = 0;
}

intIndex1 = dblData2.ToString().IndexOf('.');
intIndex1++;
if (dblData2.ToString().Length >= intIndex1 + intLength)
if (intIndex1 == 0 && intLength == 0)
strValue = dblData2.ToString();
else
strValue = dblData2.ToString().Substring(0, intIndex1 + intLength);
else
strValue = dblData2.ToString();

if (strValue!="")
if (strValue.Substring(strValue.Length - 1, 1) == ".")
strValue = strValue.Substring(0, strValue.Length - 1);

dblData3 = Convert.ToDouble(strValue);
dblData4 = Convert.ToDouble(Convert.ToDecimal(dblData2) - Convert.ToDecimal(dblData3));
}

[redmudwashcompanion]

我上面这个函数，各种情况都测过了

[caoshuming_500]

字符串处理，一直比较头痛。学习一下

[dongyangmoney]

 double a = 1.11;
            double b = 100.11;

            double qian = 0.0;
            double hou = 0.0;

            string str = string.Empty;
           

            int length = 0;
            int aindex = a.ToString().IndexOf('.');
            if (aindex == -1)
            {
                length = -1;
            }
            else
            {
                length = a.ToString().Length - aindex - 1;
            }


            int bindex = b.ToString().IndexOf('.');

            if (bindex == -1)
            {
                qian = double.Parse(b.ToString());
            }
            else
            {
                if (bindex + length + 1 > b.ToString().Length)
                {
                    qian = double.Parse(b.ToString());
                }
                else
                {
                    qian = double.Parse(b.ToString().Substring(0, bindex + length + 1));
                    str = "0.0";
                    for (int i = 1; i < b.ToString().Length - b.ToString().IndexOf('.') - 1; i++)
                    {
                        str += "0";
                    }
                    hou = double.Parse((b - qian).ToString(str));

                }
            }

[dongyangmoney]

我自己写出来了但是感觉这个 if else 太多了 不效率

[@井九]

static int GetPrecise(double b)
  {
  	if (Math.Abs(b - (int)b) < 1e-10)
  		return -1;
  	int precise = 0;
  	double temp = Math.Abs(b - (int)b);
  	while((temp = temp*10) < 1 )
  	{
  		++precise;
  	}
  	return precise;
  }
  static void Main(string[] args)
  {
  	double a = 1.1;
 		double b = 100.11;
		int precisea = GetPrecise(a);
		double result1 = ((int)(b*Math.Pow(10,precisea+1)))/(double)Math.Pow(10,precisea+1);
		double result2 = b - result1;
		Console.WriteLine(result1);
		Console.WriteLine((float)result2);
	}

[哥子谭]

提醒注意一下数据精度的问题

[peng2739956]

T-bag 难道你还没被抓？？？？？？？？？