三分去逼近一個凹/凸函式，很好的思想！這題要注意的是，迴圈終止的條件是，取的這兩個點足夠近（不是x座標足夠近）。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#include <stack>

using namespace std;

const double eps = 1e-4;

double a,b,c,x,y;

double dist(double x1,double y1,double x2,double y2){
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double getY(double xx){
	return a*xx*xx+b*xx+c;
}

double calc(double xx){
	double yy = a*xx*xx+b*xx+c;
	return dist(xx,yy,x,y);
}

int main(){

	while(cin>>a>>b>>c>>x>>y){
		double l=-10000;
		double r= 10000;
		double lres,rres;
		while(dist(l,getY(l),r,getY(r))>eps){
			double d=(r-l)/3;
			double lmid=l+d;
			double rmid=r-d;
			lres=calc(lmid);
			rres=calc(rmid);
			if(lres<rres){
				r=rmid;
			}else{
				l=lmid;
			}
		}
		double ans=calc((l+r)/2);
		printf("%.3f\n",ans);
	}
	return 0;
}