C#中如何实现微信红包功能的示例代码分享

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这篇文章主要介绍了C#实现微信红包功能,使用正态分布计算红包金额,具有一定的参考价值,感兴趣的小伙伴们可以参考一下
本文实例为大家分享了C#仿微信红包功能的具体代码,供大家参考,具体内容如下

Program.cs代码:
class Program
  {
    static void Main(string[] args)
    {
      //初始化要发起的红包基础数据
      double total = 100;
      int num = 50;
      double min = 0.01;
      string temp;
      bool flag = false;
      Console.WriteLine(string.Format("是否需要自定义红包金额和数量(默认{0}元/{1}人)Y/N:", total, num));
      temp = Console.ReadLine();
      if (temp.Trim().ToLower().Equals("y") || temp.Trim().ToLower().Equals("yes"))
      {
        Console.WriteLine("请输入你要发起的红包金额:");
        do
        {
          temp = Console.ReadLine();
          flag = double.TryParse(temp, out total);
          if (!flag)
          {
            Console.WriteLine("金额必须为整数或小数,请重新输入:");
          }
        } while (!flag);
        Console.WriteLine("请输入你要发起的红包个数:");
        do
        {
          temp = Console.ReadLine();
          flag = int.TryParse(temp, out num);
          if (!flag)
          {
            Console.WriteLine("红包个数必须为整数,请重新输入:");
          }
        } while (!flag);
      }
 
      total -= min * num;
      if (total < 0)
      {
        Console.WriteLine("抱歉,你的金额不足!");
        return;
      }
 
      //产生正态分布的随机红包金额,并计算相关的金额和数量保证数据的准确性
      double average = total / num;
      double variance = 1;
      Random u1 = new Random();
      Random u2 = new Random();
      double[] nums = new double[num];
 
      for (int i = 0; i < num; i++)
      {
        double? result = total;
        if (i < num - 1 && total > 0)
        {
          do
          {
            result = Round((double)Normal(u1.NextDouble(), u2.NextDouble(), average, variance), 2);
          } while (result == null || result < 0);
          if (total > result)
          {
            total = (double)Round((total - (double)result), 2);
          }
          else
          {
            result = total;
            total = 0;
          }
        }
        else if (i == num - 1)
        {
          total = 0;
        }
        nums[i] = Math.Round(min + (double)result, 2); //浮点运算问题,这里需要四舍五入数据才正确
 
        Console.WriteLine(string.Format("第{0}个红包金额:{1}", i + 1, (min + result)));
        Console.WriteLine("剩余金额:" + ((i != num - 1 && total == 0) ? min * (num - i - 1) : total + (min * (num - i - 1))));
      }
      Console.WriteLine("最大金额:" + nums.Max());
      Console.WriteLine("最小金额:" + nums.Min());
      Console.WriteLine("总额:" + Round(nums.Sum(), 2));
      Console.WriteLine("初始方差:" + variance);
      Console.WriteLine("结果方差:" + Variance(nums));
      Console.WriteLine("按任意键退出!");
 
      Console.ReadKey();
    }
 
    /// 
    /// 产生符合正态分布的随机数
    /// 
    /// 正态分布第一个随机数
    /// 正态分布第二个随机数
    /// 正态期望(平均值)
    /// 正态标准差(Math.Sqrt(方差))
    /// 
    public static double? Normal(double u1, double u2, double averageValue, double variance)
    {
      double? result = null;
      try
      {
        result = averageValue + Math.Sqrt(variance) * Math.Sqrt((-2) * Math.Log(u1)) * Math.Sin(2 * Math.PI * u2);
      }
      catch (Exception)
      {
        result = null;
      }
 
      return result;
    }
 
    /// 
    /// 求一组数据的方差
    /// 
    /// 要求的数组
    /// 
    public static double Variance(double[] nums)
    {
      double average = nums.Sum() / nums.Length;
      double sum = 0;
      double variance = 0;
      foreach (double num in nums)
      {
        sum += Math.Pow((num - average), 2);
      }
      variance = sum / nums.Length;
 
      return variance;
    }
 
    /// 
    /// 截取小数指定小数位,且不四舍五入
    /// 
    /// 要截取的小数
    /// 截取小数后位数
    /// 
    public static double? Round(double originNum, int lastNum)
    {
      double? result = null;
      int index = originNum.ToString().IndexOf('.');
      if (index != -1)
      {
        string temp = originNum.ToString();
        result = Convert.ToDouble(temp.Substring(0, index + 1) + temp.Substring(index + 1, Math.Min(temp.Length - index - 1, lastNum)));
      }
      if (result == 0)
      {
        result = null;
      }
      else if (index == -1)
      {
        result = originNum;
      }
 
      return result;
    }
  }
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