蒙特卡洛法并行求解Pi值C++程序

以下是并行求解 Pi 值的 C++ 代码，使用蒙特卡洛方法：

#include <iostream>
#include <random>
#include <thread>
#include <future>

int main() {
    const int num_threads = 4;          // 线程数
    const int num_points = 1000000;    // 模拟点数
    int num_in_circle = 0;             // 圆内点数

    std::vector<std::future<int>> futures(num_threads);

    std::random_device rd;
    std::vector<std::mt19937> gens(num_threads, std::mt19937(rd()));
    std::vector<std::uniform_real_distribution<>> dis(num_threads, std::uniform_real_distribution<>(-1.0, 1.0));

    for (int i = 0; i < num_threads; ++i) {
        futures[i] = std::async(std::launch::async, [&num_points, &num_in_circle, &gens, &dis, i]() {
            int local_num_in_circle = 0;
            for (int j = 0; j < num_points / num_threads; ++j) {
                double x = dis[i](gens[i]);
                double y = dis[i](gens[i]);
                if (x * x + y * y <= 1) {
                    ++local_num_in_circle;
                }
            }
            return local_num_in_circle;
        });
    }

    for (auto& future : futures) {
        num_in_circle += future.get();
    }

    double pi = 4.0 * num_in_circle / num_points;
    std::cout << "Pi = " << pi << std::endl;

    return 0;
}

该程序使用 `std::async` 函数创建多个异步任务，每个任务在一个独立的线程中运行，模拟在单位正方形内随机投点，判断每个点是否在以原点为圆心、半径为 $1$ 的圆内，最后将每个线程计算得到的圆内点数相加，再根据圆内点数和总点数的比例计算 Pi 值。