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;
}