hdu 2899 凸性 二分 ／ 三分

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7230    Accepted Submission(s): 4996

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

2 100 200

 

 

Sample Output

-74.4291 -178.8534

 

二分先决条件是满足单调性，这里显然不满足不能直接二分，这道题目求导就可以解决，求出极值点，如果求导不方便的题目就得用三分搜索了。

 

#include<iostream>
#include<stdio.h>
#include<math.h>
#define exp 1e-8

using namespace std;

double y;

double suan(double x)
{
    return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x-y;
}
double binary(double low,double high)
{
    while(high-low>exp)
    {
        double mid = (low+high)/2;
        if(suan(mid)>0)
            high=mid-exp;
        else
            low=mid+exp;
    }
    return low;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        double ans = binary(0,100);
        printf("%.4f\n",6*pow(ans,7)+8*pow(ans,6)+7*pow(ans,3)+5*ans*ans-y*ans);
    }
    return 0;
}

 

三分查找

#include<iostream>
#include<stdio.h>
#include<math.h>
#define exp 1e-8

using namespace std;

double y;

double suan(double x)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;
}
double binary(double low,double high)
{
    double mid,mmid;
    while(high-low>exp)
    {
        mid = (low+high)/2;
        mmid = (high+mid)/2;
        if(suan(mid)>suan(mmid))
            low=mid+exp;
        else
            high=mmid-exp;
    }
    return (mid+mmid)/2;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        double ans = binary(0,100);
        printf("%.4f\n",suan(ans));
    }
    return 0;
}