Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9479    Accepted Submission(s): 6447

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

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求函式的最小值
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
當導數為0時就是函式的極值點
他的導數42X^6+48X^5+21X^2+10X-Y，很明顯只有一個轉折點所以在這裡取得最小值，
實際上就是求他的導數等於0時候的值。這一題用三分法也是可以的。用導數更直接。

#include<bits/stdc++.h>
using namespace std;
#define eps 1e-8
int y;
double f(double x)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double ff(double x)
{
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&y);
        double low=0,high=100,mid;
        while(high-low>eps){
            mid=(low+high)/2;
            if(ff(mid)<eps) low=mid;
            else high=mid;
        }
        printf("%.4lf\n",f(mid));
    }
    return 0;
}