algorithm - 如何在 C++ 中生成具有固定总和的随机值

Question

我想连续生成 3 个 0 到 9 范围内的随机数，它们的总和应为给定的固定数。例如，对于给定的固定总和 15，一种可能的解决方案是 (3, 8, 4)。我怎样才能做到这一点 ？谢谢。

Answer 1

我们可以:

1. 首先生成0和1之间的随机 float a,b,c
2. 获取a,b,c的和
3. 将 a,b,c 除以 sum
4. 将a,b,c乘以给定的整数和，然后将a,b,c四舍五入到最接近的整数
5. 看看 sum(a, b, c) == given integer ?得到结果:再试一次

检查这个演示:

使用 boost 随机生成器:

#include <iostream>
#include <time.h>
#include <iomanip>
#include <boost/random.hpp>

int main()
{
    static time_t seed = time(0);
    boost::random::mt19937 RandomNumGen(seed++);
    boost::random::uniform_real_distribution<> Range(0, 1);

    int Desired_Integer = 15;
    int Rand_Max = 9;
    int Max_Itr = 100000000;
    int Count = 0;
    int SumABC[3][10] = { 0 };
    float bias = 0.5;

    float a, b, c;
    for (int Loop = 1; Loop <= Max_Itr; ++Loop)
    {
        a = Range(RandomNumGen);
        b = Range(RandomNumGen);
        c = Range(RandomNumGen);

        float Sum = a + b + c;
        a = a / Sum;
        b = b / Sum;
        c = c / Sum;

        //Round to the nearest integer;
        int aI = static_cast<int>(a * Desired_Integer + bias), bI = static_cast<int>(b * Desired_Integer + bias), cI = static_cast<int>(c * Desired_Integer + bias);
        if (aI <= Rand_Max && bI <= Rand_Max && cI <= Rand_Max && aI + bI + cI == Desired_Integer)
        {
            SumABC[0][aI]++;
            SumABC[1][bI]++;
            SumABC[2][cI]++;

            Count++;
        }
    }

    int PaddingWidth = 10;
    std::cout << "\n" << Count << " in " << Max_Itr << " loops get desired outcome. \nDistribution of a,b,c: \n";
    std::cout << "Number" << std::setw(PaddingWidth) << "a" << std::setw(PaddingWidth) << "b" << std::setw(PaddingWidth) << "c" << std::endl;
    for (int i = 0; i < 10; i++)
    {
        std::cout 
            << i << std::setw(PaddingWidth + 8) 
            << std::setprecision(4) << 100.0 * SumABC[0][i] / (float)Count << std::setw(PaddingWidth) 
            << std::setprecision(4) << 100.0 * SumABC[1][i] / (float)Count << std::setw(PaddingWidth)
            << std::setprecision(4) << 100.0 * SumABC[2][i] / (float)Count << std::endl;
    }

    std::cout << "\n\n";

    system("pause");
    return 0;
}

测试效率: