這篇文章主要為大家相許介紹了輕松實(shí)現(xiàn)微信搶紅包的ASP.NET代碼，具有一定的參考價(jià)值，感興趣的小伙伴們可以參考一下

群里都在玩搶紅包,搶了再發(fā)，發(fā)了再搶?zhuān)?jiǎn)直是無(wú)聊，程序員感興趣是的如何實(shí)現(xiàn)，這里簡(jiǎn)單說(shuō)說(shuō)實(shí)現(xiàn)思路，附上dome，代碼有點(diǎn)low，好在是實(shí)現(xiàn)了，具體內(nèi)容如下?

正文

100塊發(fā)30個(gè)紅包

50塊發(fā)13個(gè)紅包

1塊發(fā)10個(gè)紅包

發(fā)紅包需要滿(mǎn)足以下幾個(gè)條件

1.總金額不變
2.每個(gè)紅包都必須有錢(qián)
3.盡量的均勻點(diǎn),不然搶紅包沒(méi)什么意思了?

實(shí)現(xiàn)思路1.首先要確定最小單位,這里是精確到分,我這里以int類(lèi)型進(jìn)行計(jì)算,得出的結(jié)果也全是int類(lèi)型
2.數(shù)據(jù)均勻,這里以? 13.余數(shù)處理,當(dāng)數(shù)據(jù)平均給20個(gè)紅包,肯定跟總金額有所偏差,這時(shí)候我們就要多退少補(bǔ),如果大于總金額,讓紅包金額多的去減少(總金額/總?cè)藬?shù)),如果還有剩余,就讓第二多的去減少,直到補(bǔ)完這個(gè)空缺為止,反之亦然.
4.打亂順序,細(xì)心的園友可能發(fā)現(xiàn),紅包是有順序規(guī)律的,我們必須簡(jiǎn)單的寫(xiě)個(gè)function去打亂它.　　

demo

/// <summary>
?///?搶紅包
?///?</summary>
?///?<param name="money"></param>
?///?<param name="num"></param>
?///?<returns></returns>
?public?List<int>?qhb(int?money,?int?num)
?{
??int?min?=?1;
??int?pjs?=?money?/?num;
??List<int>?list?=?new?List<int>();
??Random?rnd?=?new?Random();
??for?(int?i?=?0;?i?< num; i++)
  {
  int max = (money / (num - i)) * 2;
  int s = rnd.Next(min, max);
  s = checkmoney(s, min, max);
  list.Add(s);
  money -= s;
  }
  //前面數(shù)據(jù)都是合理的,結(jié)尾必須多退少補(bǔ)
  list = checklist(money, pjs, list);
  return list;
 }

/// <summary>
?///?余數(shù)處理
?///?</summary>
?///?<param name="money"></param>
?///?<param name="list"></param>
?///?<returns></returns>
?public?List<int>?checklist(int?money,?int?pjs,?List<int>?list)
?{
??if?(money?!=?0)
??{
??if?(money?>?0)
??{

???List<int>?list_order?=?maopao(list);
???//給最小
???if?(money?/?pjs?==?0)
???{???
???list_order[0]?+=?money;
???}
???else
???{
???for?(int?i?=?0;?i?<?money?/?pjs?+?1;?i++)
???{
????if?(i?==?money?/?pjs)
????{
????list_order[i]?+=?(money?-?(money?/?pjs)?*?pjs);
????}
????else
????{
????list_order[i]?+=?pjs;
????}

???}
???}
???return?list_order;
??}
??else
??{

???List<int>?list_order?=?maopao(list,"desc");
???//給最大
???if?(money?/?pjs?==?0)
???{

???list_order[0]?+=?money;
???}
???else
???{
???for?(int?i?=?0;?i?< -(money / pjs) + 1; i++)
   {
    if (i == -(money / pjs))
    {
    list_order[i] += (money - (money / pjs) * pjs);
    }
    else
    {
    list_order[i] -= pjs;
    }

   }
   }
   return list_order;
  }
  }
  return list;
 }

 /// <summary>
??///?冒泡排序
??///?</summary>
??///?<param name="list"></param>
??///?<param name="order"></param>
??///?<returns></returns>
??public?List<int>?maopao(List<int>?list,string?order?=?"asc")
??{
??if?(order?!=?"asc")
??{
???for?(int?i?=?0;?i?< list.Count; i++)
   {
   for (int j = 0; j < list.Count - i - 1; j++)
   {
    if (list[j + 1] >?list[j])
???{
????int?temp?=?list[j];
????list[j]?=?list[j?+?1];
????list[j?+?1]?=?temp;
?
????}
???}
???}
??}
??else?
??{
???for?(int?i?=?0;?i?< list.Count; i++)
   {
   for (int j = 0; j < list.Count-i-1;j++)
   {
    if (list[j+1] < list[j])
    {
    int temp = list[j];
    list[j] = list[j+1];
    list[j+1] = temp;

   }
   }
  }
  }
  return list;
 }

/// 

?///?打亂順序
??///?
??///?
??///?
?public?List?suiji(List?list)?
??{
??Random?rnd?=?new?Random();
??for?(int?i?=?0;?i?
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