# GDUT_寒假训练题解报告_专题I_K题 个人题解报告

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

## 思路步骤：

### 2.第一步说起来短，实际上画图步骤要得出这个结论需要一定的图像处理能力，但是也不算特别难。

------------二分和三分不同，二分定的是：这一边满足条件，那么答案一定在另一边，然后不断缩小范围，使满足条件的最大/最小点得出，但是三分不同。
------------三分用于，在一个范围内寻找最小值或者最大值，三分的思想可以如此来说明：

——>分四份取三份是什么意思呢？这两个点设为mid1，和mid2
mid1=(l+r)/2,mid2=(r+mid)/2或者(l+mid)/2，
——>分三份就容易理解了：mid1=(l2+r)/3;;;mid2=(l+r2)/3

1.如果答案在mid1左边，那么F(mid1)<F(mid2)
2.如果答案在mid1和mid2之间，那么有F(mid1)<F(mid2)或者F(mid1)>F(mid2)
3.如果答案在mid2右边，有F(mi1)>F(mid2)

``````#define ULL unsigned long long
using namespace std;
typedef struct element
{
double a,b,c;
}ele;

ele all[10010];

int t,n;
double f(int i,double x)
{//fi(x)的求值函数
return all[i].a*x*x+all[i].b*x+all[i].c;
}
double F(double x)
{
double _MAX=f(0,x);
for(int time=1;time<n;time++)
{
if(f(time,x)>_MAX)_MAX=f(time,x);
}
return _MAX;
}

int main()
{
scanf("%d",&t);
for(int time=0;time<t;time++)
{
scanf("%d",&n);
for(int time1=0;time1<n;time1++)
{
scanf("%lf %lf %lf",&all[time1].a,&all[time1].b,&all[time1].c);
}
double l=0;
double r=1000;
while(r-l>1e-9)
{
double mid1=(l*2+r)/3;
double mid2=(l+r*2)/3;

if(F(mid1)<F(mid2))r=mid2;
else l=mid1;
}
printf("%.4f\n",F(l));
}
return 0;
}
``````