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.
InputThe 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)
OutputJust the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
Sample Input
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.
2 100 200Sample Output
-74.4291 -178.8534
题意:输入一个Y,求方程的最小解。
题解:用三分法,注意精度。
AC代码为;
#include<iostream>
#include<cmath>
#include<cmath>
#include<algorithm>
using namespace std;
double f(double x,double y)
{
return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
int main()
{
int T;
double y;
cin>>T;
while(T--)
{
cin>>y;
double t1,t2,mid1,mid2;
t1=0,t2=100;
for(int i=0;i<200;i++)
{
mid1=(t1+t2)/2;
mid2=(mid1+t2)/2;
if(f(mid1,y)>f(mid2,y))
{
t1=mid1;
}
else
{
t2=mid2;
}
}
printf("%.4lf
",f(mid1,y));
}
return 0;
}