随机抽取n个和等于100的整数

Draw n random integers whose sum is equal to 100

伪代码示例:

Random function: (1 to 5, values less than 100, sum must be equal to 100 when all random numbers are added).

结果示例:

Number 1 = 35
Number 2 = 15
Number 3 = 10
Number 4 = 20
Number 5 = 20

你的问题不是很明确。有许多可能的解决方案,具有不同的属性。

这是我想到的第一个:通过选择 n[=,将区间 [0, 1] 细分为 n 个部分30=] − [0, 1] 中随机的 1 个点。然后将其缩放到 [0, A] 并使用这些子区间的长度作为您的 n 随机数,总和 A.

function GetRandomNumbers(ACount: Integer; const ASum: Double): TArray<Double>;
var
  Itvs: TArray<Double>;
  i: Integer;
begin

  if ACount < 1 then
    raise Exception.Create('GetRandomNumbers: Invalid parameters.');

  // Create a subdivision of [0, 1]
  SetLength(Itvs, ACount + 1);
  Itvs[0] := 0;
  for i := 1 to ACount - 1 do
    Itvs[i] := Random;
  Itvs[ACount] := 1;
  TArray.Sort<Double>(Itvs);

  SetLength(Result, ACount);
  for i := 0 to ACount - 1 do
    Result[i] := ASum * (Itvs[i + 1] - Itvs[i]);

end;

例如,这可能会给

16.7746451916173
7.29391833301634
22.1434036735445
3.25182809028775
50.5362047115341

对于 n = 5 和 A = 100。

这使用了类似于泛型的现代 Delphi 技术,但总体思路应该足够清晰,因此您可以在 Delphi 7 中实现它并使用您喜欢的任何排序方法。另外,我将把它作为练习来制作 GetRandomNumbers.

的整数版本

仅使用整数和 Fisher-Yates 随机播放:

program cont3;
{$APPTYPE CONSOLE}
{$R *.res}
uses
  System.SysUtils;
const
  SummandsCount = 5;
  WantedSum = 100;
var
  i, j, t, Cnt, WhereToInsert: integer;
  JustNaturalNumbers: array[1..WantedSum] of Integer;
  DividingPoints: array[0..SummandsCount] of Integer;
begin
  Randomize;
  Cnt := 1;
  DividingPoints[0] := 0;
  DividingPoints[SummandsCount] := 100;
  for i := 1 to WantedSum - 1 do
    JustNaturalNumbers[i] := i;
  for i := WantedSum - 1 downto WantedSum - SummandsCount + 1 do begin
    j := 1 + Random(i);
    WhereToInsert := Cnt;
    while (WhereToInsert > 1) and (JustNaturalNumbers[j] < DividingPoints[WhereToInsert-1]) do begin
      Dec(WhereToInsert);
      DividingPoints[WhereToInsert + 1] := DividingPoints[WhereToInsert]
    end;
    DividingPoints[WhereToInsert] := JustNaturalNumbers[j];
    JustNaturalNumbers[j] := JustNaturalNumbers[i];
    Inc(Cnt);
  end;
  t := 0;
  for i := 1 to SummandsCount do begin
    Writeln(DividingPoints[i] - DividingPoints[i-1]);
    t := t + (DividingPoints[i] - DividingPoints[i-1]);
  end;
    Writeln('Sum = ', t);
  Readln;
end.

输出示例:

22
4
7
18
49
Sum = 100