分配锦标赛系统的奖品 [英] Distribute prizes for a tournament system

问题描述

我正在寻找一种在x单位之间分配数字的方法.我什至不知道该怎么说，所以我举一个例子:

I'm looking for a way to distribute a number across x units. I don't even know how to put this words so I'll give an example:

有一场比赛的总奖金为1000美元.我希望排名前20位的获胜者/参赛者能从中获胜.
我需要一个数学算法/公式来在各个球员中进行分配，并使我有能力控制某些其他分配因素.

There's a tournament in which the total prize is $1000. I want that the top 20 winners/entrants will win something out of it.
I need a mathematical algorithm/formula which will distibute it across those players, and which gives me the power to control certain other factors of the distribution.

例如，一个因素是我希望排名第一的获胜者将获得$ 300.排名第二的优胜者所占的比例较小.总分配必须给每个人一些东西，直到排名前20位的获胜者(最后一位)将至少获得X $.
X $是我要控制的另一个因素.

A factor for example is that I want the top #1 winner will get $300. The top #2 winner will get smaller percentage of it. The total distribution must give everyone something, until the top #20 winner (the last one) which will get at least X$.
X$ is another factor I want to control.

有什么主意吗?这个问题有名字吗(那是什么名字)?有代码示例吗?

Any idea? Does this problem has a name (and what's that name is)? Any code example?

编辑#1-我的第一个建议:

#include <conio.h>
#include <vector>

#define TOTAL                       100
#define WINNERS                     15
#define FIRST_WINNER_PERCENTAGE     0.30

void distribute_1(::std::vector<double> * const prizes)
{
    prizes->clear();

    double total = TOTAL;
    double winning_percentage = FIRST_WINNER_PERCENTAGE;
    double slope = 0.5;
    int winners = WINNERS;

    double winning = 0;
    for(int i = 0; i < winners; i++, total -= winning, winning_percentage /= 2)
    {
        winning = total * winning_percentage;
        prizes->push_back(winning);
    }
}
void distribute_2(::std::vector<double> * const prizes)
{
    prizes->clear();

    double total = TOTAL;
    double winning_percentage = FIRST_WINNER_PERCENTAGE;
    double slope = 0.5;
    int winners = WINNERS;

    double winning = 0;
    for(int i = 0; i < winners; i++, total -= winning/*, winning_percentage /= 2*/)
    {
        winning = total * winning_percentage;
        prizes->push_back(winning);
    }
}
void distribute_3(::std::vector<double> * const prizes)
{
    prizes->clear();

    double total = TOTAL;
    double winning_percentage = FIRST_WINNER_PERCENTAGE;
    double slope = 0.0005;
    int winners = WINNERS;

    double winning = 0;
    for(int i = 0; i < winners; i++, total -= winning, winning_percentage -= slope)
    {
        winning = total * winning_percentage;
        prizes->push_back(winning);
    }
}
void distribute_4(::std::vector<double> * const prizes)
{
    prizes->clear();

    double total = TOTAL;
    double winning_percentage = FIRST_WINNER_PERCENTAGE;
    double slope = 1 / WINNERS;
    int winners = WINNERS;

    double winning = 0;
    for(int i = 0; i < winners; i++, total -= winning, winning_percentage -= slope)
    {
        winning = total * winning_percentage;
        prizes->push_back(winning);
    }
}

void main()
{
    ::std::vector<double> prizes;

    distribute_1(&prizes);
    distribute_2(&prizes);
    distribute_3(&prizes);
    distribute_4(&prizes);

    double total_granted = 0;
    for(int i = 0; i < WINNERS; i++)
    {
        total_granted += prizes[i];
        printf("%lf\n", prizes[i]);
    }
    printf("-\n%lf\n", total_granted);

    _getch();
}

这是我所能达到的.例如，如果您将"WINNERS"设置为5，则该算法未达到"TOTAL"量(在此示例中为100)，甚至还没有达到(我总共有83)，问题就出在这个问题上.

This is as far as I could reach. The issue with this one is for example, if that if you set 'WINNERS' to 5 for example, the algorithm doesn't reach the 'TOTAL' amount (100 in this example) or even closer (I get a total of 83).

克里斯蒂的解决方案:

#include <conio.h>
#include<iostream>
//using arithmetic progression
using namespace std;
int i;
float ratio;
float first_prize;
float s;
int main()
{
    float money=1000;
    const int total_prizes =        10;
    float last_prize =              99;
    float prizes[total_prizes+1];

    /**/first_prize=2*money/total_prizes-last_prize; //last member of the progresion
    ratio=(first_prize-last_prize)/(total_prizes-1);
    prizes[total_prizes]=last_prize;
    for(i=total_prizes-1;i>=1;i--){
       prizes[i]=prizes[i+1]+ratio;
       money-=prizes[i];
    }
    for(i=1;i<=total_prizes;i++){
        printf("%d) %.2f\n",i,prizes[i]);
        s+=prizes[i];
    }
    printf("TOTAL SUM:%.2f\n",s);
    printf("Ratio: %.2f", ratio);
    _getch();
}

推荐答案

现在是上午1:15，我正在解决数学:)).
使用算术级数.
我使用了所有定义，因此您可以轻松地对其进行更改.

It's 1:15 AM here and I'm solving maths :)).
Using arithmetic progression.
I made all using defines so you can easily change them.

#include<iostream>
//using arithmetic progression
using namespace std;
FILE *g=fopen("output.out","w");
#define last_prize 10
#define total_prizes 20
int i;
float prizes[total_prizes+1];
float money=1000;
float ratio;
float first_prize;
float s;
//a1=last_prize
//an=first_prize
int main(){
 first_prize=2*money/total_prizes+last_prize; //last member of the progresion
 ratio=(first_prize-last_prize)/(total_prizes-1);
 prizes[total_prizes]=last_prize;
    for(i=total_prizes-1;i>=1;i--)
       prizes[i]=prizes[i+1]+ratio;
 for(i=1;i<=total_prizes;i++){
  fprintf(g,"%d) %.2f\n",i,prizes[i]);
  s+=prizes[i];
 }
 fprintf(g,"TOTAL SUM:%.2f",s);
return 0;
}

输出:

1) 90.00
2) 85.79
3) 81.58
4) 77.37
5) 73.16
6) 68.95
7) 64.74
8) 60.53
9) 56.32
10) 52.11
11) 47.89
12) 43.68
13) 39.47
14) 35.26
15) 31.05
16) 26.84
17) 22.63
18) 18.42
19) 14.21
20) 10.00
TOTAL SUM:1000.00

如您所见，它们的总和为1000.00 $:D

As you can see they sum up to exactly 1000.00$ :D

其他结果:
输入:

Other results:
INPUT:

#define last_prize 30
#define total_prizes 5

输出:

1) 370.00
2) 285.00
3) 200.00
4) 115.00
5) 30.00
TOTAL SUM:1000.00

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