HDOJ 2899 Strange fuction(二分，三分)

寻找函数最小值

最新推荐文章于 2019-04-17 23:23:03 发布

原创最新推荐文章于 2019-04-17 23:23:03 发布 · 365 阅读

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本文介绍了一种通过二分法和三分法寻找特定多项式函数在指定区间内的最小值的方法，并提供了完整的C语言实现代码。

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

//二分之凸性
#include <stdio.h>
#include <math.h>
const double eps = 1e-8;
double y;
double cal(double x)    //f(x)的求导式
{
    return 42.0*pow(x,6.0)+48.0*pow(x,5.0)+21.0*pow(x,2.0)+10.0*x;
}
double ans(double x)    //f(x)原式
{
    return 6.0*pow(x,7.0)+8.0*pow(x,6.0)+7.0*pow(x,3.0)+5.0*pow(x,2.0)-y*x;
}
int main()
{
    int T;
    double f,l,mid;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf",&y);
        /*
        if(cal(100.0)-y<=0.0)       //这点可以不用判断
        {
            printf("%.4lf\n",ans(100.0));
            continue;
        }

        if(cal(0)-y>0.0)
        {
           printf("0.0000\n");
           continue;
        }
        */
        f=0.0,l=100.0;
        while(l-f>eps)    //二分就是这么神奇，你值得拥有
        {
            mid=(f+l)/2.0;
            if(cal(mid)-y<0.0)
            {
                f=mid;
            }
            else
            {
                l=mid;
            }
        }
        printf("%.4lf\n",ans(mid));
    }
    return 0;
}

//3分法
#include<stdio.h>
#include<math.h>
double math(double X,double Y)
{
    return 6*pow(X,7)+8*pow(X,6)+7*pow(X,3)+5*pow(X,2)-Y*X;
}
int main()
{
    int T;
    double x,y,Y,a,b,c,z,x1,y1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf",&Y);
        x=0;
        y=100;
        while(y-x>0.000001)   //3分法
        {
            x1=x+(y-x)/3;
            y1=x+(y-x)/3+(y-x)/3;
            a=math(x1,Y);
            b=math(y1,Y);
            if(a<b)
                y=y1;
            else
                x=x1;
        }
        printf("%.4lf\n",math(y,Y));
    }
    return 0;
}