生成n个随机数,其总和为m,所有数字应大于零

时间:2014-03-13 13:57:56

标签: java random

我想生成9个非零随机数,其总和为250。 我试过下面的代码,它给了我9个随机数,但有些数字为零。

 public void n_random()
{
  Random r = new Random();
ArrayList<Integer> load = new ArrayList<Integer>();
    int temp = 0;
    int sum = 0;
    for (int i = 1; i <= 9; i++) {
        if (!(i == 9)) {
            temp = r.nextInt(250 - sum);
            System.out.println("Temp " + (i) + "    " + temp);
            load.add(temp);
            sum += temp;

        } else {
            int last = (250 - sum);
            load.add(last);
            sum += last;
        }
    }

    System.out.println("Random arraylist " + load);
    System.out.println("Sum is "+ sum);

}

我的错误在哪里或我应该在哪里改进我的代码或任何其他解决方案?

8 个答案:

答案 0 :(得分:10)

我建议使用:

temp = r.nextInt((250 - sum) / (9 - i)) + 1;

这将确保:

  • 每个号码都是严格正面的
  • 在达到第9个号码之前,您不会使用完整的“250津贴”

然而,结果的分布可能有偏差。

示例输出:

  

随机arraylist [18,28,22,19,3,53,37,49,21]

说明:

  • (250 - sum)是要达到250的金额,所以你不想过去那个
  • / (9 - i)如果您的金额已经达到200(需要50多个)并且您还有5个要去,请确保下一个随机数不超过10,为接下来的4个抽奖留出一些空间
  • + 1以防止0

可能提供更好分布的替代方案是获取随机数并缩放它们以获得所需的总和。示例实现:

public static void n_random(int targetSum, int numberOfDraws) {
    Random r = new Random();
    List<Integer> load = new ArrayList<>();

    //random numbers
    int sum = 0;
    for (int i = 0; i < numberOfDraws; i++) {
        int next = r.nextInt(targetSum) + 1;
        load.add(next);
        sum += next;
    }

    //scale to the desired target sum
    double scale = 1d * targetSum / sum;
    sum = 0;
    for (int i = 0; i < numberOfDraws; i++) {
        load.set(i, (int) (load.get(i) * scale));
        sum += load.get(i);
    }

    //take rounding issues into account
    while(sum++ < targetSum) {
        int i = r.nextInt(numberOfDraws);
        load.set(i, load.get(i) + 1);
    }

    System.out.println("Random arraylist " + load);
    System.out.println("Sum is "+ (sum - 1));
}

答案 1 :(得分:5)

  

生成n个随机数,其总和为m,所有数字应大于零

以下基本上是您要实现的目标。在这里,它是用Perl编写的,因为我不太熟悉Java,但它应该很容易翻译。

use strict;
use warnings;
use feature qw( say );

use List::Util qw( shuffle );

my $m = 250;
my $n = 9;
my @nums;
while ($n--) {
   my $x = int(rand($m-$n))+1;  # Gen int in 1..($m-$n) inclusive.
   push @nums, $x;
   $m -= $x;
}

say join ', ', shuffle @nums;   # shuffle reorders if that matters.

你的方法的问题是你会得到很多小数字。五个样本运行,数字按升序排列:

  • 1,1,1,1,2,3,6,50,185
  • 1,1,1,1,2,3,4,13,224
  • 1,1,1,1,1,3,8,11,223
  • 1,1,1,1,2,4,19,103,118
  • 2,2,9,11,11,19,19,68,109

更好的方法可能是采用N个随机数,然后对它们进行缩放,使其总和达到M.实现:

use strict;
use warnings;
use feature qw( say );

use List::Util qw( sum );

my $m = 250;
my $n = 9;

# Generate $n numbers between 0 (incl) and 1 (excl).
my @nums;
for (1..$n) {
   push @nums, rand();
}

# We subtract $n because we'll be adding one to each number later.
my $factor = ($m-$n) / sum(@nums);

for my $i (0..$#nums) {
   $nums[$i] = int($nums[$i] * $factor) + 1;
}

# Handle loss of fractional component.
my $fudge = $m - sum(@nums);
for (1..$fudge) {
   # Adds one to a random number.
   ++$nums[rand(@nums)];
}

say join('+', @nums), '=', sum(@nums);

五个样本运行:

32+32+23+42+29+32+29+20+11=250
31+18+25+16+11+41+37+56+15=250
21+15+40+46+22+40+32+1+33=250
34+24+18+29+45+30+19+29+22=250
3+45+20+6+3+25+18+65+65=250

答案 2 :(得分:3)

您的代码行:

r.nextInt(250 - sum);

...将生成从0包含)到250 - sum已排除)的伪随机数。

Random.nextIntAPI

我不会尝试在此处解决您的所有问题,但只需将1添加到上面的表达式即可保证它永远不会返回0

所有剩余的改编虽然取决于你:)

例如,如果250 - sum - 1评估为否定,那么您将抛出IllegalArgumentException

答案 3 :(得分:0)

我没有看到您检查的位置,以确保排除零。在将其插入数组之前添加一个检查。

此代码中有太多“神奇数字”以适合我。

答案 4 :(得分:0)

Random.nextInt(n)的调用将返回0到n-1之间的整数

尝试temp = r.nextInt(250 - sum) + 1;,看看是否能解决您的问题。

答案 5 :(得分:0)

这是一种方法,可以避免(大多数)魔术数字并提供合适的数字分布,尽管所有数字都会比其他可能的解决方案更小。

public static void n_random(int count, int finalSum)
{
    Random r = new Random();
    int numbers[] = new int[count];
    int sum = 0;
    for (int i = 0; i < count - 1; i++)
    {
        numbers[i] = r.nextInt((finalSum - sum) / 2) + 1;
        sum += numbers[i];
    }
    numbers[count - 1] = finalSum - sum;

    StringBuilder numbersString = new StringBuilder("Random number list: ");
    for (int i = 0; i < count; i++)
        numbersString.append(numbers[i] + " ");
    System.out.println(numbersString);
}

public static void main(String[] args)
{
    n_random(9, 250);
}

答案 6 :(得分:0)

这是javascript替代

function getRandomNos(m,n) {

    var nums=[];
    var i;

    for (i = 0;i<n;i++) {
       nums[i] = Math.random();
    }
    var factor = (m-n) / nums.reduce(add, 0);
    for (i = 0;i<nums.length;i++) {
       nums[i] = parseInt(nums[i] * factor) + 1;
    }

    var fudge = m - nums.reduce(add, 0);

    for (i=0;i<fudge;i++) {
       nums[i] = nums[i] + 1;
    }
    console.log(nums);
    console.log(nums.reduce(add, 0));

}

function add(a, b) {
    return a + b;
}

答案 7 :(得分:0)

//Perl code translated to Java
//Its working and no 0's !!!!

import java.util.*;
import java.util.stream.*;enter code here

public class MyClass {
    public static void main(String args[]) {
     int numberOfDraws = 17;
     int targetSum = 40;
     


Random r = new Random();


List<Integer> load = new ArrayList<>();



int sum = 0;
for (int i = 0; i < numberOfDraws; i++) {
        int next = r.nextInt(targetSum) + 1;
        load.add(next);
        sum += next;
        System.out.println("Arraylist first loop " + load.get(i));
    }
    
    
double factor = (((double)targetSum)-((double)numberOfDraws)) / ((double)sum);
System.out.println("Factor value: " + factor);

int newSum =0;
for (int i = 0; i < numberOfDraws; i++) {
    load.set(i, (int) ((load.get(i) * factor)) + 1);
    newSum += load.get(i);
    
    System.out.println("Arraylist second loop " + load.get(i));
}



int fudge = targetSum - newSum;
for (int i = 0; i < fudge; i++) {

   
   int y = r.nextInt(numberOfDraws);
        load.set(i, load.get(i) + 1);
   
}

System.out.println("Random arraylist " + load);

    }
    }