绘制n个等于100的随机整数 [英] Draw n random integers whose sum is equal to 100
本文介绍了绘制n个等于100的随机整数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
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
推荐答案
仅使用整数和Fisher-Yates随机播放:
Using only integer numbers and Fisher-Yates shuffle:
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
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