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;
}