Luogu 10013

题目分析：

• 本题是求最小值，三分时，令 $l_{mid}=l+(r-l)/3,r_{mid}=r-(r-l)/3,如果f\left(r_{mid}\right)\>=f\left(l_{mid}\right),就舍弃l_{mid}右边那一截，令r=r_{mid}$lmid​=l+(r−l)/3,rmid​=r−(r−l)/3,如果f(rmid​)>=f(lmid​),就舍弃lmid​右边那一截，令r=rmid​
• 精度要开小点

Code:

#include <bits/stdc++.h>
using namespace std;
#define eps 1e-9
#define maxn 100010

int T,n;
double a[maxn],b[maxn],c[maxn];

inline double AKNOI_(double x) {
	double maxd=-2139062143;
	for(int i=1;i<=n;++i) {
		maxd=max(maxd,a[i]*x*x+b[i]*x+c[i]);
	}
	return maxd;
}

void readda_() {
	scanf("%d",&T);
	for(int AKIOI=1;AKIOI<=T;++AKIOI) {
		scanf("%d",&n);
		for(int i=1;i<=n;++i) {
			scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);
		}
		double ans,l=0,r=1000,mid,l_mid,r_mid;
		while(r-l>eps) {
			mid=(r-l)/3;
			l_mid=l+mid;
			r_mid=r-mid;
			if(AKNOI_(l_mid)<AKNOI_(r_mid)) {
				ans=l_mid;
				r=r_mid;
			}
			else l=l_mid;
		}
		printf("%.4lf\n",AKNOI_(ans));
	}
}

int main() {
	freopen("a.txt","r",stdin);
	readda_();
	return 0;
}