hdu Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 269    Accepted Submission(s): 233

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

分析：方法一、对导数二分求的导数为0点；方法二、直接对函数三分求最值。

二分代码：

#include<cstdio>
#include<cmath>
#define eps 1e-15
double f(double x,double y)
{
    return 6 * pow(x,7)+8*pow(x,6)+7*x*x*x+5*x*x-y*x;
}
double g(double x)
{
    return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x;
}
int main()
{
    int T;
    double y,a,b,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf",&y);
        if(g(0.0)>=y)
            printf("%.4f\n",f(0.0,y));
        else if(g(100.0)<=y)
            printf("%.4f\n",f(100.0,y));
        else
        {
            a=0.0;
            b=100.0;
            while(b-a>eps)
            {
                m=(a+b)/2;
                if(g(m)>y)
                    b=m;
                else
                    a=m;
            }
            printf("%.4f\n",f(m,y));
        }
    }
    return 0;
}

三分代码：

#include<cstdio>
#include<cmath>
#define eps 1e-15
double y;

double f(double x) {
    return 6 * pow(x, 7) + 8 * pow(x, 6) + 7 * x * x * x + 5 * x * x - y*x;
}

int main() {
    int T;
    double a, b, c, d;
    scanf("%d", &T);
    while (T--) {
        scanf("%lf", &y);
        a = 0.0;
        b = 100.0;
        while (b - a > eps) {
            c = (a + b) / 2;
            d = (b + c) / 2;
            if (f(c) < f(d))
                b = d;
            else
                a = c;
        }
        printf("%.4f\n", f(a));
    }
    return 0;
}