hdu 2899 Strange fuction（用三分即可 不用求导）

转载亲请注明出处：http://blog.csdn.net/u012860063/article/details/21318849

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2899

----------------------------------------------------------------------------------------------------------------------------------------------------------

欢迎光临天资小屋：http://user.qzone.qq.com/593830943/main

----------------------------------------------------------------------------------------------------------------------------------------------------------

首先简介一下三分思想：

大家都知道二分搜索算法作为分治中最常见的方法，适用于单调函数，逼近求解某点的值。但当函数具有凹凸性时，二分搜索就无法使用了（当然可以先排序），这里介绍一种新的方法，三分查找算法，适应的函数图形如下：

算法思想

1. 与二分查找算法类似，先取整个区间的中间值mid

mid = (left + right) / 2;

2. 再去右侧区间的中间值midmid，从而把区间分成三个小区间

midmid = (mid + right) / 2;

3. 如果mid比midmid更靠近最值点，我们就舍弃右区间(midmid + 1 ~ end),否则我们就舍弃左区间(left ~ mid - 1).

比较mid与midmid谁更靠近最值，只需要确定mid所在的函数值与midmid所在的函数值的大小。当最值为最大值时，mid与midmid中较大的那个自然更为靠近最值。最值为最小值时同理

if (cal(mid) < cal(midmid)) 
	left = mid;
else
	right = midmid;

4. 重复1,2,3,直到找到最值点

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2712    Accepted Submission(s): 2010

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
   
   
   
   

 

Author

Redow

 

Recommend

lcy

题意：即求最小值！

代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define EPS 1e-7     //开始的时候设置成五位 一直wa
using namespace std;
int y;
double Calc(double x)
{
    return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0)-y*x;
}
double Solve(void)
{
    double Left, Right;
    double mid, midmid;
    double mid_value, midmid_value;
    Left = 0; Right = 100;
    while (Right - Left > EPS )
    {
        mid = (Left + Right) / 2;
        midmid = (mid + Right) / 2;
        mid_value = Calc(mid);
        midmid_value = Calc(midmid);
        if (mid_value < midmid_value) 
			Right = midmid;
        else 
			Left = mid;
    }
    return Left;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&y);
        printf("%.4lf\n",Calc( Solve()));
    }
    return 0;
}

赞！不懂请评论提出！