hdu 2899 Strange fuction (二分法)

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2278    Accepted Submission(s): 1697

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

 

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//0MS    224K    590 B    G++
/*
    
    题意：
        给出一关系式： 
            F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) 
        和y的值，求x在范围(0,100)内时F(x)能取得的最小值
    
    二分法：
        很容易可以看出函数F(x)在(0,100)内是先减后增的，故用二分法求其倒数的零值点，然后用零值
    再代入原式即可求出最值   
         
*/
#include<stdio.h>
#define e 1e-7
double fac(double a,double b)
{
    return 6*a*a*a*a*a*a*a+8*a*a*a*a*a*a+7*a*a*a+5*a*a-b*a;    
} 
double fac0(double a,double b)
{
    return 42*a*a*a*a*a*a+48*a*a*a*a*a+21*a*a+10*a-b;       
}
int main(void)
{
    int t;
    double n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&n);
        double l=0;
        double r=100;
        while(r-l>e){
            double mid=(l+r)/2;
            if(fac0(mid,n)>0) r=mid;
            else l=mid;
        }
        printf("%.4lf
",fac(r,n));
    }
    return 0;
}