关于三分的学习

借鉴优秀的博客关于三分的解释
练习题目
曲线
题解

#include<cstdio>
#include<algorithm>
#include<string>
using namespace std;

int n;
double a[100005], b[100005], c[100005];
double check(double x)
{
	double maxx = a[1]*x*x + b[1]*x + c[1];
	for(int i = 2; i <= n; i++)
		maxx = max(maxx, a[i]*x*x + b[i]*x + c[i] );
	return maxx;
 } 
 
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i = 1; i <= n; i++)
			scanf("%lf %lf %lf", &a[i], &b[i], &c[i]);
		double l = 0, r = 1000;
		while(r-l >= 1e-9)
		{
			double mid1 = l + (r-l) / 3.0 , mid2 = r - (r-l) / 3.0;
			if(check(mid1) > check(mid2)) 
				l = mid1;
			else r = mid2;
		 } 
		 printf("%.4lf\n",check(l));
	}
	return 0;
}

灯泡 超级棒的题解
题解

#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;

double D, H, h;
int t;

double check(double x)
{
	double l1 = D - x;
	double l2 = H - (H-h) * D/x;
	return l1 + l2;
}

int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf %lf %lf",&H, &h, &D);
		double l = (H-h) * D/H, r = D;
		while(r - l >= 1e-6)
		{
			double mid1 = l + (r-l) / 3.0, mid2 = r - (r-l) / 3.0;
			if(check(mid1) > check(mid2)) 
				r = mid2;
			else l = mid1;
		}
		printf("%.3lf\n",check(l));
	}
	return 0;
}

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