通过减少量将 x 分成 y 部分

Question

如果我有 1000 美元（可变），我想把这笔钱分成 20 个（可变）人，但不是平均分配给每个人，而是想给第一个人更多，然后第二人称等

所以第 20 个人得到的最少，第 5 个人得到第 5 的最多。

我将如何实现这一目标？

谢谢

编辑：

公式：

int people = 20;
float prize = 1000;

float k = (2 * prize) / ((people) * (people - 1));
float sum = 0;

for (int i = 1; i < people; ++i)
{
    var personsPrize = i * k;
    sum += personsPrize;
    Console.WriteLine(personsPrize);
}
Console.WriteLine("sum = " + sum);

Answer 1

公式正确，需要稍加修改。

1. 不要将 float 转换为 int，数据会丢失！

2. 在for中进行时，从第一人称到n-1人

   int people = 20;
   float prize = 1000;
   
   float k = (2 * prize) / ((people) * (people - 1));
   float sum = 0;
   
   for (int i = 1; i < people; ++i)
   {
       var personsPrize = i * k;
       sum += personsPrize;
       Console.WriteLine(personsPrize);
   }
   Console.WriteLine("sum = " + sum);
   

Answer 2

给Ensure prize pool doesn't award tied participants less than participants who scored worse添加答案

两件事

1. 不正确，当分配价格从第一人到第n-1人时，一个人没有中奖，正确的是从第一人到第n >

2. 用 K 标记的有效 basePrize 是

   k = (2 * 奖品) / ((人) * (人 + 1))

现在回答另一个问题：

int totalPeople = 20;
float prize = 5000;

List<int> peopleScores = new List<int>();
Random r = new Random();
for (int i = 0; i < totalPeople; ++i)
{
    peopleScores.Add(r.Next(0, 100));
}

var totalPeopleWithScore = peopleScores.Where(x => x > 0).Count();

var groupedScores = peopleScores
    .Where(x => x > 0)
    .ToList()
    .GroupBy(x => x)
    .Select(grp => new
    {
        score = grp.Key,
        peopleScores = grp.ToList()
    })
    .OrderBy(g => g.score)
    .ToList();

var groupCount = groupedScores.Select(x => new { count = x.peopleScores.Count() }).Select(z => z.count).Count();

float basePrizeRate = 2 * prize / totalPeopleWithScore / (totalPeopleWithScore + 1);
Console.WriteLine("Base Price rate: " + basePrizeRate);
float sum = 0;
var leaderboardPosition = 0;
var totalWinners = 0;
foreach (var positionScore in groupedScores)
{
    var countWinner = positionScore.peopleScores.Count();
    Console.WriteLine();
    Console.WriteLine($"On leaderboard position : {groupedScores.Count() - leaderboardPosition} are {countWinner} winners with score: {positionScore.score}");

    float positionPrizePool = 0;
    for (int i = 1; i <= positionScore.peopleScores.Count(); ++i)
    {
        totalWinners++;
        positionPrizePool += totalWinners * basePrizeRate;
    }
    Console.WriteLine("Prize Pool " + positionPrizePool);
    var personPoolPrize = positionPrizePool / positionScore.peopleScores.Count();
    foreach (var x in positionScore.peopleScores)
    {
        Console.WriteLine($"Winner {totalWinners} won: {personPoolPrize}");
        sum += personPoolPrize;
    }
    leaderboardPosition++;
}
Console.WriteLine();

Console.WriteLine("Total Prize: " + sum);
Console.WriteLine("Total Winners: " + totalPeopleWithScore);

Answer 3

你欠我一杯啤酒，其实是一大杯啤酒！！

使用高斯分布 (N*(N+1))/2 你不能比竞争对手的排名有微小的差距，奖金线性增加https://mathbitsnotebook.com/Algebra2/Sequences/SSGauss.html。

你需要的是呈指数增长或以固定比率增长的东西，我想用数学方法解决你的问题，所以我将使用几何级数https://mathbitsnotebook.com/Algebra2/Sequences/SSGeometric.html

我不会在这个之后添加任何其他答案，您必须建立自己的奖池系统！

对于几何分布公式为：

rankPrize = ((1 - 分布因子) / (1 - 分布因子 ^ 获胜者）* distributionFactor ^（排名 - 1））* 奖金

在哪里 distributionFactor - 介于 0 和 1 之间

玩转分配因子并为您的系统选择正确的值，我选择了 0.8，因为奖品之间的差距不是那么大！

我们开始，小提琴链接：https://dotnetfiddle.net/qmJnYd 玩转和实现：

static void GaussDistributionPricePool()
        {
            Console.WriteLine("________________________________________________________");
            Console.WriteLine("Begin Gauss distribution price pool");

            int totalPeople = 20;
            float prize = 5000;

            List<int> peopleScores = new List<int>();
            Random r = new Random();
            for (int i = 0; i < totalPeople; ++i)
            {
                peopleScores.Add(r.Next(0, 100));
            }

            var totalPeopleWithScore = peopleScores.Where(x => x > 0).Count();

            var groupedScores = peopleScores
                .Where(x => x > 0)
                .ToList()
                .GroupBy(x => x)
                .Select(grp => new
                {
                    score = grp.Key,
                    peopleScores = grp.ToList()
                })
                .OrderBy(g => g.score)
                .ToList();

            var groupCount = groupedScores.Select(x => new { count = x.peopleScores.Count() }).Select(z => z.count).Count();
            // Gauss formula. (N*(N+1))/2
            // https://mathbitsnotebook.com/Algebra2/Sequences/SSGauss.html
            float basePrizeRate = 2 * prize / totalPeopleWithScore / (totalPeopleWithScore + 1);
            Console.WriteLine("Base Price rate: " + basePrizeRate);
            float sum = 0;
            var leaderboardRank = 0;
            var totalWinners = 0;
            foreach (var positionScore in groupedScores)
            {
                var countWinner = positionScore.peopleScores.Count();
                Console.WriteLine();
                Console.WriteLine("On leaderboard rank : " + (groupedScores.Count() - leaderboardRank) + " are " + countWinner + " winners with score: " + positionScore.score);

                float positionPrizePool = 0;
                for (int i = 1; i <= positionScore.peopleScores.Count(); ++i)
                {
                    totalWinners++;
                    positionPrizePool += totalWinners * basePrizeRate;
                }
                Console.WriteLine("Prize Pool " + positionPrizePool);
                var personPoolPrize = positionPrizePool / positionScore.peopleScores.Count();
                foreach (var x in positionScore.peopleScores)
                {
                    Console.WriteLine("Winner " + totalWinners + " won: " + personPoolPrize);
                    sum += personPoolPrize;
                }
                leaderboardRank++;
            }
            Console.WriteLine();

            Console.WriteLine("Total Prize: " + sum);
            Console.WriteLine("Total Winners: " + totalPeopleWithScore);

            Console.WriteLine("End Gauss distribution price pool");
            Console.WriteLine("________________________________________________________");
        }