阿里算法内推笔试题

#include <iostream>
#include <vector>
#include <numeric>
#include<limits>
#include <cstdio>
#include <math.h>
#include <time.h>
#include <cstdlib>
#define pi 3.1415926
#define NUM 10000
using namespace std;

/** 请完成下面这个函数，实现题目要求的功能 **/
 /** 当然，你也可以不按照这个模板来作答，完全按照自己的想法来 ^-^  **/
double cal(double x,double mul,double sigma)          //计算概率密度函数
{
    double t=sqrt(2*pi);
    double t1=-(x-mul)*(x-mul)/(2*sigma*sigma);
    double f=t*exp(t1);
    return f;
}
bool inside(double x,double y)                   //判断是否在内部
{
    double a=(x*x+y*y-1)*(x*x+y*y-1)*(x*x+y*y-1)-x*x*y*y*y;
    if(a<=0)
        return true;
    else
        return false;
}

double randgass(double minvalue, double maxvalue,double miu, double sigma){//生成正态随机数
    double x,y,dscope;
    do{
        x=minvalue+(maxvalue-minvalue)*rand()/(RAND_MAX+1.0);
        y=cal(miu,miu,sigma);
        dscope=y*rand()/(RAND_MAX+1.0);
    }while(y<dscope);
    return x;
}

double leartCurve(double mu1, double sigma1, double mu2, double sigma2) {
    double xrange=mu1-3*sigma1;
    double xsplit=6*sigma1;
    double yrange=mu2-3*sigma2;
    double ysplit=6*sigma2;
    double res=0;
    double xp=0,yp=0;
    int num=0;
    srand((unsigned)time(NULL));
    for(int i=0;i<NUM;i++)
    {
        xp=randgass(xrange,xrange+xsplit,mu1,sigma1);
        yp=randgass(yrange,yrange+ysplit,mu2,sigma2);
        if (inside(xp,yp)) num++;
    }
    res=(double)num/NUM;
    return res;
}

int main() {
    double res;

    double _mu1;
    cin >> _mu1;
    //cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');

    double _sigma1;
    cin >> _sigma1;
    //cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');

    double _mu2;
    cin >> _mu2;
    //cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');

    double _sigma2;
    cin >> _sigma2;
    //cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');


    res = leartCurve(_mu1, _sigma1, _mu2, _sigma2);
    cout << res << endl;

    return 0;

}