C - Strange fuction

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

先求导数，可以看出在0到100区间里面是单峰函数，主要是用二分法求导数为0的地方，注意小数位数，我的代码弄得比较复杂了。

代码如下：

#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#define EXP 1e-9
using namespace std;

double y;

double f(double x)
{
return 42 * pow(x,6) + 48 * pow(x,5) + 21 * pow(x,2) + 10 * x - y;
}

double bs()
{
double lo = 0, hi = 100;
double mi, ans = hi + 1;
while(lo + EXP < hi)
{
mi = (lo + hi) / 2.0;
if(f(mi) < 0)
{
lo = mi;
if(ans = hi + 1)
ans = mi;
max(mi, ans);
}
else
{
hi = mi;
min(mi, ans);
}
//cout << lo << " " << hi << endl;
}
//cout << ans << endl;
if(ans == 100 + 1)
return -1;
return ans;
}

int main()
{
int t;
cin >> t;
while(t--)
{
cin >> y;
double temp = bs();
//cout << temp << endl;
//cout << fixed << setprecision(5) << (6 * pow(temp,7) + 8 * pow(temp,6) + 7 * pow(temp,3) + 5 * pow(temp,2) - y * temp) * 10000 << endl;
if(temp != -1.0)
{
if((6 * pow(temp,7) + 8 * pow(temp,6) + 7 * pow(temp,3) + 5 * pow(temp,2) - y * temp) * 10000 >= 0)
cout << fixed << setprecision(4) << (long long)((6 * pow(temp,7) + 8 * pow(temp,6) + 7 * pow(temp,3) + 5 * pow(temp,2) - y * temp) * 10000 + 0.5) / 10000.0 << endl;
else
cout << fixed << setprecision(4) << (long long)((6 * pow(temp,7) + 8 * pow(temp,6) + 7 * pow(temp,3) + 5 * pow(temp,2) - y * temp) * 10000 - 0.5) / 10000.0 << endl;
}
else
{
//cout << "?" << endl;
if(f(0) >= 0 && f(100) >= 0)
cout << fixed << setprecision(4) << (double)0 << endl;
else
cout << fixed << setprecision(4) << (long long)((6 * pow(100,7) + 8 * pow(100,6) + 7 * pow(100,3) + 5 * pow(100,2) - y * 100) * 10000 - 0.5) / 10000.0  << endl;
}
}
return 0;
}