题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2899
Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 350 Accepted Submission(s): 275
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.
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
思路：
这个题是叫我们求最小值，那么很明显对于函数求最值，最简单的方法就是求导
我们队f(x)求导之后得到：f'(x)=42*x^6+48*x^5+21*x*x+10x-y
很明显，当f'(x)=0时取得最值，然后由于定义域x属于[0,100]那么可以推断，f'(x)=0时
f(x)有最小值。
我们可以利用二分算法去计算f'(x)=0时x为多少（注意f'(x)是单调递增的）
计算出x之后再代入原函数解出最小值即可。
我的代码：
#include<stdio.h>
#include<math.h>
double df(double x)
{
return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x;
}
double f(double x,double y)
{
return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;
}
int main()
{
int t,i;
double y,left,right,mid;
scanf("%d",&t);
while(t--)
{
scanf("%lf",&y);
left=0,right=100;
for(i=1;i<=100;i++)
{
mid=(left+right)/2;
if(df(mid)>y)
right=mid;
else
left=mid;
}
printf("%.4lf\n",f(mid,y));
}
return 0;
}