Strange fuction 2899 （二分+数学求导）

Strange fuction

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

Total Submission(s): 4717    Accepted Submission(s): 3383


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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)
 

[align=left]Output[/align]
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
 

[align=left]Sample Input[/align]

2
100
200

 

[align=left]Sample Output[/align]

-74.4291
-178.8534//一个简单的二分题，但得要先判断一下导数的正负来确定函数的输出//比较简单（一遍AC）

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
double fun(double x)
{
return 42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x;
}
double f(double x,double y)
{
return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
}
int main()
{
int t;
double y;
scanf("%d",&t);
while(t--)
{
scanf("%lf",&y);
if(fun(0)>y)
printf("0.0000\n");
else if(fun(100)<y)
printf("%.4lf\n",f(100.0,y));
else if(fun(0)<y&&fun(100)>y)
{
double l=0,r=100,mid;
while(r-l>1e-10)
{
mid=(l+r)/2;
if(fun(mid)-y>1e-6)
r=mid;
else
l=mid;
}
printf("%.4lf\n",f(mid,y));
}
}
return 0;
}