C# が受信時刻と今日の時差を計算する方法の例の共有

/// <summary>
       /// 计算传入的时间距离今天的时间差
       /// </summary>
       /// <param name="dt"></param>
       /// <param name="yy"></param>
       /// <param name="mm"></param>
       /// <param name="dd"></param>
       public void GetCriminalYX(DateTime dt, out int yy, out int mm, out int dd)
       {
           DateTime now = DateTime.Now;
           yy = mm = dd = 0;
           if (dt.Year > 9000 || dt.Year == 1900)
           {
               return;
           }
           if (dt <= now)
           {
               return;
           }
           StringBuilder str = new StringBuilder();
           int dt_Y = dt.Year;
           int dt_M = dt.Month;
           int dt_D = dt.Day;
           int now_Y = DateTime.Now.Year;
           int now_M = DateTime.Now.Month;
           int now_D = DateTime.Now.Day;
           yy = dt_Y - now_Y;
           mm = dt_M - now_M;
           dd = 0;
int dt_M_SY = 0;
           if (dt_D < now_D)
           {
               mm -= 1;
               dt_M_SY = dt_M - 1;
               if (dt_M_SY == 0)
               {
                   dt_M_SY = 12;
               }
               if (dt_M_SY == 2)
               {
                   dt_M_SY = dt_Y % 4 == 0 ? 29 : 28;
               }
               else
               {
                   dt_M_SY = dt_M_SY == 2 || dt_M_SY == 4 || dt_M_SY == 6 || dt_M_SY == 9 || dt_M_SY == 11 ? 30 : 31;
               }
               dt_D += dt_M_SY;
           }
           dd = dt_D - now_D;
           if (mm < 0)
           {
               yy -= 1;
               mm += 12;
           }
       }