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抽奖算法

网上找的抽奖代码都不满意,自己动手丰衣足食,有需要的可以直接拿去用,不收版权费的。

/// <summary>/// 抽奖/// </summary>public class Prize{    /// <summary>    /// 奖品关键字    /// </summary>    public string Key { get; set; }    /// <summary>    /// 权重/数量    /// </summary>    public int Poll { get; set; }    /// <summary>    /// 中奖区间    /// </summary>    class Area    {        /// <summary>        /// 奖品关键字        /// </summary>        public string Key { get; set; }        /// <summary>        /// 开始索引位置        /// </summary>        public int Start { get; set; }        /// <summary>        /// 截止索引位置        /// </summary>        public int Over { get; set; }    }    /// <summary>    /// 随机种子    /// </summary>    static Random Rand = new Random((int)DateTime.Now.Ticks);    /// <summary>    /// 轮盘抽奖,权重值(在轮盘中占的面积大小)为中奖几率    /// </summary>    /// <param name="prizeList">礼品列表(如果不是百分百中奖则轮空需要加入到列表里面)</param>    /// <returns></returns>    public static string Roulette(List<Prize> prizeList)    {        if (prizeList == null || prizeList.Count == 0) return string.Empty;        if (prizeList.Any(x => x.Poll < 1)) throw new ArgumentOutOfRangeException("poll权重值不能小于1");        if (prizeList.Count == 1) return prizeList[0].Key; //只有一种礼品        Int32 total = prizeList.Sum(x => x.Poll); //权重和                       if (total > 1000) throw new ArgumentOutOfRangeException("poll权重和不能大于1000"); //数组存储空间的限制。最多一千种奖品(及每种奖品的权重值都是1)        List<int> speed = new List<int>(); //随机种子        for (int i = 0; i < total; i++) speed.Add(i);        int pos = 0;        Dictionary<int, string> box = new Dictionary<int, string>();        foreach (Prize p in prizeList)        {            for (int c = 0; c < p.Poll; c++) //权重越大所占的面积份数就越多            {                pos = Prize.Rand.Next(speed.Count); //取随机种子坐标                box[speed[pos]] = p.Key; //乱序 礼品放入索引是speed[pos]的箱子里面                speed.RemoveAt(pos); //移除已抽取的箱子索引号            }        }        return box[Prize.Rand.Next(total)];    }    /// <summary>    /// 奖盒抽奖,每个参与者对应一个奖盒,多少人参与就有多少奖盒    /// </summary>    /// <param name="prizeList">礼品列表</param>    /// <param name="peopleCount">参与人数</param>    /// <returns></returns>    public static string LunkyBox(List<Prize> prizeList, int peopleCount)    {        if (prizeList == null || prizeList.Count == 0) return string.Empty;        if (prizeList.Any(x => x.Poll < 1)) throw new ArgumentOutOfRangeException("poll礼品数量不能小于1个");        if (peopleCount < 1) throw new ArgumentOutOfRangeException("参数人数不能小于1人");        if (prizeList.Count == 1 && peopleCount <= prizeList[0].Poll) return prizeList[0].Key; //只有一种礼品且礼品数量大于等于参与人数        int pos = 0;        List<Area> box = new List<Area>();        foreach (Prize p in prizeList)        {            box.Add(new Area() { Key = p.Key, Start = pos, Over = pos + p.Poll }); //把礼品放入奖盒区间            pos = pos + p.Poll;        }        int total = prizeList.Sum(x => x.Poll); //礼品总数        int speed = Math.Max(total, peopleCount); //取礼品总数和参数总人数中的最大值        pos = Prize.Rand.Next(speed);        Area a = box.FirstOrDefault(x => pos >= x.Start && pos < x.Over); //查找索引在奖盒中对应礼品的位置        return a == null ? string.Empty : a.Key;    }}/*List<Prize> prizes = new List<Prize>();prizes.Add(new Prize() { Key = "电脑", Poll = 1 });prizes.Add(new Prize() { Key = "机柜", Poll = 2 });prizes.Add(new Prize() { Key = "鼠标", Poll = 3 });                        string lp1 = Prize.LunkyBox(prizes, 6);Console.WriteLine(lp1);prizes.Add(new Prize() { Key = "谢谢惠顾", Poll = 5 });string lp2 = Prize.Roulette(prizes);      Console.WriteLine(lp2);*/

 

抽奖算法