【二分解方程】hdu 2899 Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7719    Accepted Submission(s): 5268

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

2100200

 

Sample Output

-74.4291-178.8534

 
///求函数最小值，求导后导数单调增，直接求其零点即是最小值点
///AC代码

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;double f(double x, double y){    return 6 * pow(x, 7) + 8 * pow(x, 6) + 7 * pow(x, 3) + 5 * x * x - y * x;}double ff(double x, double y){    return 42 * pow(x, 6) + 48 * pow(x, 5) + 21 * x * x + 10 * x - y;}int main(){    int n;    cin >> n;    for (int i = 0; i < n; i++)    {        double y;        cin >> y;        double left = 0, right = 100, mid = 50;        while (right - left > 1e-6)        {            mid = (left + right) / 2;            if (ff(mid, y) > 0)            {                right = mid;            }            else if (ff(mid, y) < 0)            {                left = mid;            }            else            {                break;            }        }        cout << fixed << setprecision(4) << f(mid, y) << endl;    }    return 0;}