Strange fuction Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5933 Accepted Submission(s): 4194 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 Recommend
lcy //三分二分都可以，导数单调，求导后二分值 AC代码
#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define MX 10000 + 50
using namespace std;
double y; double f(double x) { //求导
return *pow(x,)+*pow(x,)+*x*x+*x-y; // F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
} double F(double x) {
return * pow(x,)+*pow(x,)+*x*x*x+*x*x-y*x;// F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
} int main() {
int T;
while(cin>>T) {
for(int qq=; qq<T; qq++) {
cin>>y;
if(f()>) {
printf("%.4f\n",F());
} else if(f()<) {
printf("%.4f\n",F());
} else {
double l=; // 0 l r 100
double r=; // -------------------------
double mid;
while(r-l>1e-) {
mid=(r+l)/2.0;
if(f(mid)>) {
r=mid;
} else {
l=mid;
}
}
printf("%.4f\n",F(mid));
}
}
}
return ;
}