Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 732 Accepted Submission(s): 600 

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

2100200

 

Sample Output

-74.4291-178.8534

求这个函数的最小值F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)

二分三分应该是都可以出来的，控制住精度，这里用的循环

#include<cmath>#include <iostream>#include<stdio.h>using namespace std;double y;double gg(double x){    return  6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;}int main(){    int t,i;    double n;    cin>>t;    while(t--)    {        cin>>y;        double low=0,high=100,mid;        double temp1=0,temp2=0;        for (i=1;i<=100;i++)                     {            temp1=(low+high)/2;            temp2=(temp1+high)/2;                               if (gg(temp1)<gg(temp2))                high=temp2;            else                low=temp1;        }        printf("%.4lf\n",gg(temp1));    }    return 0;}