文章标题HDU2899：Strange fuction？（二分+精度）

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
题意：求解f（x）在0到100之间的最小值
分析：有题意可知可对f（x）求导，判断导函数与y值的大小关系进行二分。
代码：

#include<iostream>#include<cstdio>#include<cstring>#include<math.h>#include<algorithm>using namespace std;double F(double x,double y){    return (6*x*x*x*x*x*x+8*x*x*x*x*x+7*x*x+5*x-y)*x; }long double F1(double x){    return 42*x*x*x*x*x*x+x*x*x*x*x*48+21*x*x+10*x;}int main (){    double y;    int total;    scanf ("%d",&total);    while(total--){        scanf ("%lf",&y);        double lo=0.0,hi=100.0;        double mi=50.0;        int k=200;        while (k--){            if (F1(mi)<y)lo=mi;            else if (F1(mi)>y)hi=mi;            mi=(hi+lo)/2.0;        }        printf ("%.4lf\n",F(mi,y));    }    return 0;}

注：所示代码通过控制循环次数的方法得到精度；除此之外，自身也可以定义一个精度，但得防止精度过高导致死循环。