hdoj 2899 Strange fuction (三分搜索 二分搜索)

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6878    Accepted Submission(s): 4811

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

 

一、求函数最值，先求导成单调函数，二分判断导数为0的值；

二、函数不单调，但是有凸性，三分方法求最值。

code ：

二分：

#include <iostream>#include<cstdio>#include<cmath>using namespace std;double y;double Fx(double x){    return 6.0*pow(x,7)+8.0*pow(x,6)+7.0*pow(x,3)+5.0*x*x-y*x;}double Fx1(double x){    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;}int main(){    int n;    scanf("%d",&n);    while(n--)    {        scanf("%lf",&y);        if(Fx1(100)<=0)            printf("%.4lf\n",Fx(100));        else if(Fx1(0)>=0)            printf("%.4lf\n",Fx(0));        else        {            double l,r,mid;            l=0;r=100;            while(r-l>=1e-10)            {                mid=(l+r)/2;                if(Fx1(mid)<0)                    l=mid;                else                    r=mid;            }            printf("%.4lf\n",Fx(mid));        }    }    return 0;}

三分：

#include <iostream>#include<cstdio>#include<cmath>using namespace std;double y;double Fx(double x){    return 6.0*pow(x,7)+8.0*pow(x,6)+7.0*pow(x,3)+5.0*x*x-y*x;}int main(){    int n;    scanf("%d",&n);    while(n--)    {        scanf("%lf",&y);        double l,r,mid,midmid;        l=0.0;r=100.0;        while(r-l>=1e-10)        {            mid=(l+r)/2;            midmid=(mid+r)/2;            if(Fx(mid)<Fx(midmid))                r=midmid;            else                l=mid;        }        printf("%.4lf\n",Fx(mid));    }    return 0;}