答案是3/4

分析不太好分析

代码：

#include <iostream>
#include <vector>
#include <random>
#include <algorithm>
#include <cmath>

using namespace std;

// 计算两点间弧长（较短的那段）
double getArcLength(double a, double b) {
    double diff = abs(a - b);
    return min(diff, 1.0 - diff);
}

// 判断是否为钝角三角形
bool isObtuseTriangle(double x1, double x2, double x3) {
    // 按顺序排列三个点
    // vector<double> points = {x1, x2, x3};
    vector<double> points;
    points.push_back(x1);
    points.push_back(x2);
    points.push_back(x3);
    sort(points.begin(), points.end());
    
    // 计算三段弧长（顺时针方向的较短弧）
    double arc1 = getArcLength(points[0], points[1]); // AB
    double arc2 = getArcLength(points[1], points[2]); // BC
    double arc3 = getArcLength(points[2], points[0]); // CA
    
    // 计算对角的弧（不包含该顶点的弧）
    double oppArc1 = arc2 + arc3; // 对角 BCA 的弧 BC + CA
    double oppArc2 = arc1 + arc3; // 对角 BAC 的弧 AB + CA
    double oppArc3 = arc1 + arc2; // 对角 ABC 的弧 AB + BC
    
    // 只要有两段弧之和小于 0.5，就形成钝角三角形
    return (oppArc1 < 0.5 || oppArc2 < 0.5 || oppArc3 < 0.5);
}

int main() {
    // 设置随机数生成器
    random_device rd;
    mt19937 gen(rd());
    uniform_real_distribution<> dis(0.0, 1.0);
    
    int trials = 1000000; // 模拟次数
    int obtuseCount = 0;
    
    // 模拟过程
    for (int i = 0; i < trials; i++) {
        // 随机生成三个点
        double x1 = dis(gen);
        double x2 = dis(gen);
        double x3 = dis(gen);
        
        // 检查是否形成钝角三角形
        if (isObtuseTriangle(x1, x2, x3)) {
            obtuseCount++;
        }
    }
    
    // 计算概率
    double probability = static_cast<double>(obtuseCount) / trials;
    cout << "模拟次数: " << trials << endl;
    cout << "形成钝角三角形的次数: " << obtuseCount << endl;
    cout << "概率: " << probability << endl;
    cout << "理论值: 0.75" << endl;
    
    return 0;
}