hdu 2899 Strange fuction(二分+数学)

题目来源：hdu 2899 Strange fuction

Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4829 Accepted Submission(s): 3441

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

题目大意：
给出一个函数：F(x)=6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100);现在输入一个y值.(0 < Y <1e10)，求此函数的最小值。
题目分析：
观察函数，我们可知，F(x)一阶导数为F’(x)=42 * x^6+48*x^5+21*x^2+10*x-y;而F’(x)中若不看y,则前面的式子为单增函数，当输入y值后，我们便可把y当做常数来处理，当y值大于前面式子的最大值，即F’(100)<0时，原函数F(x)单调递减，此时F(100)为最小值；而当y不大于前面的式子，即存在y使得F’(x)可以==0时，则我们可以知道，x取F’(x)==0的这个x值时，原函数F(x)取得最小值，在此处便需要使用二分来查找这个x值。
AC代码：

#include<stdio.h>#include<math.h>double F(double x)      //将原函数分解为F(x)-y*x，此处求的是分解后的F(x)值 {    return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0);}double f(double x)      //分解后的F'(x)的值 {    return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*pow(x,1.0);}int main(){    int n;    double y;    while(scanf("%d",&n)!=EOF)    {        while(n--)        {            scanf("%lf",&y);            if(y>=F(100))           //此时原函数的导数不存在等于0的情况，即小于零                 printf("%.4lf\n",F(100)-y*100);            else            {                double mid,left=0,right=100;                while(right-left>1e-8)  //使用二分进行查找使得分解后的F(x)的导数等于0的x                 {                    mid=(left+right)/2;                    if(f(mid)<=y)                        left=mid;                    else                        right=mid;                }                printf("%.4lf\n",F(mid)-y*mid);            }        }    }    return 0;}