ACM : HDU 2899 Strange fuction 解题报告 -二分、三分

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 ;
 }