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.
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
感觉一开始完全被唬到了 听完凯神说的才知道三分其实和二分没什么本质差别
#include<cstdio> #include<iostream> #include<algorithm> using namespace std;//F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) double pow(double x,int n) { double ans=1.0; for(int i=1;i<=n;i++) { ans*=x; } return ans; } double fun(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 n; double x,y; double l,r; double midl,midr; cin>>n; while(n--) { cin>>y; l=0.0;r=100.0; while(r-l>=0.0000001)// 这里必须足够小 否则误差较大 { midl=l+(r-l)/3; midr=l+2*(r-l)/3; if(fun(midl,y)<fun(midr,y)) r=midr; else l=midl; } printf("%.4lf ",min(fun(l,y),fun(r,y))); } return 0; }