简单抽奖算法

最新推荐文章于 2025-05-02 14:55:48 发布

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最新推荐文章于 2025-05-02 14:55:48 发布

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原文链接：http://www.cnblogs.com/shaddock/p/6186946.html

        /// <summary>
        /// 获取抽奖结果
        /// </summary>
        /// <param name="prob">奖品列表</param>
        /// <returns>中奖实体</returns>
        private LotteryGift Get(List<LotteryGift> prob)
        {
            LotteryGift result = null;
            int n = (int)(prob.Sum(t => t.GiftProbability) * 1000);           //计算概率总和，放大1000倍
            Random r = rnd;
            decimal x = (decimal)r.Next(0, n) / 1000;       //随机生成0~概率总和的数字

            for (int i = 0; i < prob.Count(); i++)
            {
                decimal pre = prob.Take(i).Sum(p => p.GiftProbability);         //区间下界
                decimal next = prob.Take(i + 1).Sum(s => s.GiftProbability);    //区间上界
                if (x >= pre && x < next)               //如果在该区间范围内，就返回结果退出循环
                {
                    result = prob[i];
                    break;
                }
            }
            return result;
        }

转载于:https://www.cnblogs.com/shaddock/p/6186946.html