HDU 2899 Strange fuction——分治+数学

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2899

Strange fuction

Problem Description

Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
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

2 100 200

 

Sample Output

-74.4291 -178.8534

 

Author

Redow

 

Recommend

lcy

 

题解：

　　题目意思就是求题目中的目标函数的在[0,100]内的最小值。

 

　　次数太高了，怎么办？？？

 

　　求个导：f(x)'=42*x^6+48*x^5+21*x^2+10*x-y，这样就把y与x分开来了，而这东西在[0,100]上是单增的（显而易见），既然具有单调性，二分答案即可了。

 

　　但还有另一个做法，我们求导之后发现它的导函数在该区间上是单调的，那么从另外一个角度来说，原函数在该区间上是有凸性的，对于一格有凸性的的函数求最值，我们就可以用三分来做了。

 

　　这道题即可以用来练二分又可以练三分，还能练数学，虽然是道水题，却不失为道好题………………

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

double get(double x,double y)
{
    return 6.0*x*x*x*x*x*x*x+8.0*x*x*x*x*x*x+7.0*x*x*x+5.0*x*x-y*x;
}

double get2(double x,double y)
{
    return 42.0*x*x*x*x*x*x+48.0*x*x*x*x*x+21.0*x*x+10.0*x-y;
}

void way1()
{
    int t;
    scanf("%d",&t);
    for (int a=1;a<=t;a++)
    {
        double y;
        scanf("%lf",&y);
        double r=100.0,l=0.0;
        while (r-l>=1e-6)
        {
            double lm=(r-l)/3.0+l;
            double rm=r-lm+l;
            if (get(lm,y)<get(rm,y)) r=rm;
            else l=lm;
        }
        printf("%.4lf\n",get(r,y));
    }
}

void way2()
{
    int t;
    scanf("%d",&t);
    for (int a=1;a<=t;a++)
    {
        double y;
        scanf("%lf",&y);
        double r=100.0,l=0.0;
        while (r-l>=1e-6)
        {
            double m=(l+r)/2.0;
            if (get2(m,y)<0) l=m;
            else r=m;
        }
        printf("%.4lf\n",get(r,y));
    }
}

int main()
{
    //way1();//三分
    way2();//求导二分

    return 0;
}