（二分搜索 ）Strange fuction -- HDU -- 2899

链接：

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

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4802    Accepted Submission(s): 3427

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

 

F'(x) = 42 * x^6+48*x^5+21*x^2+10*x-y

 

42 * x^6+48*x^5+21*x^2+10*x = y

x 要是 double 类型的， 左边要尽可能接近右边的 y 值

再带入F(x)中

F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x

 

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

#define  N 150
#define  D (1e-6)

double f(double x)
{
    return 42*pow(x, 6.0) + 48*pow(x, 5.0) + 21*pow(x, 2.0) + 10*x;
}

double F(double x)
{
    return 6*pow(x, 7.0) + 8*pow(x, 6.0) + 7*pow(x, 3.0) + 5*pow(x, 2.0);
}

int main()
{

    int t;
    scanf("%d", &t);

    while(t--)
    {
        int y;

        scanf("%d", &y);

        if(y>=f(100))
            printf("%.4f\n", F(100)-100*y);
        else
        {

            double L=0, R=100, m;

            while(R-L>D)
            {
                m = (L+R)/2;
                if(f(m)<=y) L=m;
                else R=m;
            }

            printf("%.4f\n", F(L)-L*y);
        }
    }
    return 0;
}