杭电2899Strange fuction 求导+二分

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4788    Accepted Submission(s): 3413


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

 
求导加二分，当导函数等于0时，该式取得最小值；
附ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
double x,y,a,b,i,j;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf",&y);
double left=0;
double right=100;
int m=100;
while(m--)//循环100次基本能精确到小数点后不知多少位了.
{
a=(left+right)/2;
if(42*a*a*a*a*a*a+48*a*a*a*a*a+21*a*a+10*a-y>0)//pow函数好像不能用，因为不是整形的
right=a;//当当前导函数大于0时，往 左侧查找
if(42*a*a*a*a*a*a+48*a*a*a*a*a+21*a*a+10*a-y<0)
left=a;//当当前导函数小于0时，往 右侧查找
}
//printf("%.6lf\n",a);
double ans=6*a*a*a*a*a*a*a+8*a*a*a*a*a*a+7*a*a*a+5*a*a-y*a;
printf("%.4lf\n",ans);
}
return 0;
}