1.1002

2.

Problem Description Now, here is a fuction:<br>&nbsp;&nbsp;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<br>100<br>200
Sample Output -74.4291<br>-178.8534
Author Redow

3.基本題意，對於給定的方程，找出確定範圍內的最小值x

4.F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x ，分析方程，對於x的部分求導，導數小於0遞減，倒數大於0遞增，找到導數中的拐點，拐點處函式值最小，使用二分法查詢

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;
}