2019 GDUT 新生 专题Ⅰ K题

K - 三分 HDU - 3714

题目
Josephina is a clever girl and addicted to Machine Learning recently. She
pays much attention to a method called Linear Discriminant Analysis, which
has many interesting properties.
In order to test the algorithm’s efficiency, she collects many datasets.
What’s more, each data is divided into two parts: training data and test
data. She gets the parameters of the model on training data and test the
model on test data. To her surprise, she finds each dataset’s test error curve is just a parabolic curve. A parabolic curve corresponds to a quadratic function. In mathematics, a quadratic function is a polynomial function of the form f(x) = ax2 + bx + c. The quadratic will degrade to linear function if a = 0.

It’s very easy to calculate the minimal error if there is only one test error curve. However, there are several datasets, which means Josephina will obtain many parabolic curves. Josephina wants to get the tuned parameters that make the best performance on all datasets. So she should take all error curves into account, i.e., she has to deal with many quadric functions and make a new error definition to represent the total error. Now, she focuses on the following new function’s minimum which related to multiple quadric functions. The new function F(x) is defined as follows: F(x) = max(Si(x)), i = 1…n. The domain of x is [0, 1000]. Si(x) is a quadric function. Josephina wonders the minimum of F(x). Unfortunately, it’s too hard for her to solve this problem. As a super programmer, can you help her?
Input
The input contains multiple test cases. The first line is the number of cases T (T < 100). Each case begins with a number n (n ≤ 10000). Following n lines, each line contains three integers a (0 ≤ a ≤ 100), b (|b| ≤ 5000), c (|c| ≤ 5000), which mean the corresponding coefficients of a quadratic function.
Output
For each test case, output the answer in a line. Round to 4 digits after the decimal point.
Sample Input
2
1
2 0 0
2
2 0 0
2 -4 2
Sample Output
0.0000
0.5000

题目大意及思路:给出一个或多个一元二次方程Si(x)的系数,顺序为二次项式系数,一次项系数,常数,其中a (0 ≤ a ≤ 100), b (|b| ≤ 5000), c (|c| ≤ 5000),且都为整数。
要你求F(x) = max(Si(x)),即每个x对应的F(x)取多个方程中的最大值。然后在草稿纸上画出几个一元二次函数的图,发现构成的F(x)仍为下凸函数,所以可以用三分法来求该函数最小值。
在这里插入图片描述
下面是三分过程,如果F(lmid)较小,则答案在区间(left,rmid),反之答案在区间(lmid,right).

double left=0,right=1000,rmid,lmid;
	while(right-left>p){
		lmid=(left+right)/2;
		rmid=(lmid+right)/2;
		double l_res=count(lmid);//计算F(lmid)
		double r_res=count(rmid);//计算F(rmid)
		if(l_res>r_res){
			left=lmid;
		}
		else if(l_res<r_res){
			right=rmid;
		}
		else break;
	}

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const double p=0.000000001;//误差就是得很小很小。
int a[10005],b[10005],c[10005],n;
double count(double x){
	double ans=-100000000;
	for(int i=0;i<n;i++){
		ans=max(ans,a[i]*x*x+b[i]*x+c[i]);
	}
	//printf("%f ",ans);
	return ans;
}
double f(){
	double left=0,right=1000,rmid,lmid;
	while(right-left>p){
		lmid=(left+right)/2;
		rmid=(lmid+right)/2;
		double l_res=count(lmid);
		double r_res=count(rmid);
		if(l_res>r_res){
			left=lmid;
		}
		else if(l_res<r_res){
			right=rmid;
		}
		else break;
	}
	return count((left+right)/2);
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i=0;i<n;i++){
			scanf("%d%d%d",&a[i],&b[i],&c[i]);
		}
		printf("%.4f\n",f());
	}
	return 0;
}

题目链接

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转载自blog.csdn.net/weixin_45794203/article/details/103964185