HDU 1002 Strange fuction

函数极值求解

最新推荐文章于 2020-08-17 22:11:42 发布

StormAX97

最新推荐文章于 2020-08-17 22:11:42 发布

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本文链接：https://blog.youkuaiyun.com/Eburs/article/details/51236272

本文介绍了一种通过三分法寻找给定区间内特定多项式函数最小值的方法，并提供了完整的C++实现代码。

Problem Description

Now, here is a fuction:<br>  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)<br>Can you find the minimum value when x is between 0 and 100.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

Output

Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input

2<br>100<br>200

Sample Output

-74.4291<br>-178.8534

题目大意：求函数极值

解题思路：套三分模板求极值

AC代码：

#include<iostream>
#include<fstream>
#include<cmath>
using namespace std;
double f(double x,double y)
{
double a;
a = 6 * pow(x,7)+ 8 * pow(x,6) + 7 * pow(x,3)
+ 5 * x * x - y * x;
return a;
}
int main()
{
double n;
double max,mid,midmid,min,amid,amidmid;
cin>>n;
while(cin>>n){
min = 0 ; max = 100;
while(fabs(max - min)> 1e-5)
{
mid = (min + max)/2 ;
midmid = (mid + max)/2;
amid = f(mid,n);
//cout << amid << endl ;
amidmid = f(midmid,n);
if(amid < amidmid)
max = midmid ;
else
min = mid ;

}
double c = amid;
printf("%.4f\n",c);
}
return 0;
}