练习二1002

Strange fuction
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)


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

 
题意：

在0到100之间，找到使 F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x 成立的最小值。

思路：

相当于y是个常数，求 F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)这个函数的最小值，令F' = 0，得出x，y的方程，用二分法解方程得x0(易证得x0>=0 && x0<=100)，则F'(x0) = 0,由F'' 在[0-100]上恒大于0，所以F'在[0-100]上单增，所以F'(x)<0(x<x0),F'(x)>0(x>x0),所以F(x)在x=x0处取得最小值，所以本题主要就是二分求解方程的x0，然后直接带入x0，y计算即可。

#include <cstdio>
#include <cmath>
#include <cstdlib>

const double eps = 1e-6;

double cal(double x){
return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*x;
}

double gao(double x,double y){
return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y;
}

int main(){
int t;
double low,high,y,x,res;

scanf("%d",&t);
while(t--){
scanf("%lf",&y);
low = 0.0;
high = 100.0;
while(high-low>eps){
x = (low+high)/2;
res = cal(x);
if(res<y){
low = x + 1e-8;
}else{
high = x - 1e-8;
}
}
printf("%0.4lf\n",gao(x,y));
}
return 0;
}