HDU 2899 Strange fuction 二分

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5429    Accepted Submission(s): 3886

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

 

题目大意:第一行T，接下来输入T个Y，求方程F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)最小值。

思路：求导，把导函数等于0的X值计算出来，代入原方程就是最小值。(好像有什么三分......没写，下次写下)

 

数学方法:

#include <iostream>
#include <math.h>
double y;
double vv(double x)
{
    return (6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y);
}
double v(double x)
{
    return (42*pow(x, 6)+48*pow(x, 5)+21*pow(x, 2)+10*x-y);    
}
double f(double x,double z,double y)
{
    double mid;
    while((y-z)>1e-10)
    {
    mid = (y+z)/2;
    if(v(mid)<x)
    z=mid+1e-10;
    else
    y=mid-1e-10;
    }
    return mid;
}
int main()
{
    int a, T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lf", &y);
        printf("%.4lf\n", vv(f(0,0,100)));    
    }    
    return 0;
}

 三分：

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
double f(double x, double y)
{
    return (6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y);
}
int main()
{
    int a, T, t;
    double mid1,mid2, z, y;
    cin>>T;
    while(T--)
    {
        z=0;y=100;t=50;
        cin>>a;
        while(t--){
        mid1=(z+y)/2;
        mid2=(mid1+y)/2;
        if(f(mid1,a)<f(mid2,a))
        y=mid2;
        else
        z=mid1;
        }
        printf("%.4lf\n", f(mid1,a));
    }
    return 0;
}