杭电2899--Strange fuction（二、三分）

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4603    Accepted Submission(s): 3308

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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RE: 求最值问题。

//三分法：

#include <cstdio>
#include <iostream>
using namespace std;
double f(double x, double y)
{
    return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
}
int main()
{
    int t; double m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lf", &m);
        double l_min, r_max, min = 0.0, max = 100.0;
    //    printf("%.4lf %.4lf
", min, max);
        while(max - min > 1e-8)
        {
            l_min = (2 * min + max) / 3.0;
            r_max = (min + 2 * max) / 3.0;
            if(f(l_min, m)>f(r_max, m)) min = l_min;
            else max = r_max; 
        }
    //    printf("%.4lf %.4lf
", min, max);
        printf("%.4lf
",f((min + max) / 2.0, m));
    }
    return 0;
} 

//闲来无事， 敲敲手生的二分（大同小异， 一个比较值， 一个比较导数）；

#include <cstdio>
#include <iostream>
using namespace std;
double f(double x, double y)
{
    return 42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x-y;
}
double F(double x, double y)
{
    return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x; 
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        double m, min = 0.0, max = 100.0, mid;
        scanf("%lf", &m);
        while(max - min > 1e-8)
        {
            mid = (max + min) / 2.0;
            if(f(mid, m) <= 0) min = mid;
            else max = mid;
        }
        printf("%.4lf
", F((min + max) / 2.0 , m));
    }
    return 0;
}