HDOJ 2899 Strange fuction 三分或二分

Problem Description

Now, here is afuction:<br>&nbsp;&nbsp;F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0&lt;= x &lt;=100)<br>Can you find the minimum value when x isbetween 0 and 100.

 

 

Input

The first line of the inputcontains 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

 简单三分或二分求导实现

代码实现：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
#define maxn 50005
using namespace std;
typedef long long ll;
double f(double x)
{
  return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x;
}

double f1(double x,double y)
{
  return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;
}
int main()
{
  int T;
  scanf("%d",&T);
  double n;
  while(T--)
  {
    scanf("%lf",&n);
    double l=0,r=100;
    while(l+1e-8<r)
    {
      double mid=(l+r)/2;
      
      //cout<<mid<<endl;
      if(f(mid)>=n)
        r=mid;
      else
        l=mid;
    }
    //cout<<l<<endl;
    printf("%.4lf\n",f1(l,n));
  }
  return 0;
}

#include<bits/stdc++.h>
using namespace std;
 double y;
double cal(double mid)
{
    return 6*pow(mid,7)+8*pow(mid,6)+7*pow(mid,3)+5*pow(mid,2)-y*mid;
}
int main()
{
   int T;
   cin>>T;
   int i;

   for(i=1;i<=T;i++)
   {
       cin>>y;
        double low=0,high=100,mid,midmid;
        while(high-low>1e-9)//开始连续查找函数
       {
           mid=(low+high)/2;
           midmid=(mid+high)/2;
           double cmid=cal(mid);
           double cmidmid=cal(midmid);
           if(cmid<cmidmid)
            high=midmid;
           else
            low=mid;
       }
             printf("%.4lf\n",cal(mid));

   }
    return 0;
}