搜索1002

题目大意:

给出一个方程, 6 * x^7+8*x^6+7*x^3+5*x^2-y*x,输入y,求该方程的最小值,x的变化范围为0-100

解题思路:

先对方程求导,倒数为0时的x满足方程的最小值

代码:

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
double get(double mid,double y)
{
    return 6*pow(mid,7)+8*pow(mid,6)+7*pow(mid,3)+5*mid*mid-mid*y;
}
double get1(double mid)
{
    return 42*pow(mid,6)+48*pow(mid,5)+21*pow(mid,2)+10*mid;
}
double get0(double y)
{
    double mid,low,high;
    low=0,high=100;
    while(high>low+0.0000001)
    {
        mid=(high+low)/2;
        if(get1(mid)<y)
        low=mid+0.0000001;
        else high=mid-0.0000001;
    }

    return mid;
}


int main()
{
    int T;
    double Y;
    cin>>T;
    while(T--)
    {
         cin>>Y;
         cout<<setiosflags(ios::fixed)<<setprecision(4)<<get(get0(Y),Y)<<endl;
    }
    return 0;
}
原文地址:https://www.cnblogs.com/Sikaozhe/p/5349663.html