acm 2 1002 二分查找

1.1002

2.

[align=left]Problem Description[/align]
Now, here is a fuction:<br>  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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

[align=left]Sample Input[/align]

2<br>100<br>200

[align=left]Sample Output[/align]

-74.4291<br>-178.8534

[align=left]Author[/align]
Redow

3.基本题意，对于给定的方程，找出确定范围内的最小值x

4.F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x ，分析方程，对于x的部分求导，导数小于0递减，倒数大于0递增，找到导数中的拐点，拐点处函数值最小，使用二分法查找

5.

#include<iostream>

#include<vector>

#include<algorithm>

#include<numeric>

#include<string.h>

#include<sstream>

#include<string>

#include<map>

#include<queue>

#include<iomanip>

using namespace std;

double f(double x)

{

    double l;

    l=42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x;

    return l;

}

double f1(double x)

{

    double l;

    l=6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x;

    return l;

}

int main()

{

    double n,i,k,z,l,e;

    double t,t1,t2;

    double x,y;

    cin>>n;

    while(n--)

    {

            cin>>y;

            t1=0;

            t2=100;

            while(t2-t1>0.0000001)

            {

                t=(t2+t1)/2;

                k=f(t)-y;

                if(k>0) t2=t;

                else t1=t;

            }

            z=f1(t)-y*t;

            cout<<fixed<<setprecision(4)<<z<<endl;

    }

    return 0;

}