HDU2899 Strange fuction (三分)

最新推荐文章于 2020-04-22 20:58:26 发布

TooSIMple_

最新推荐文章于 2020-04-22 20:58:26 发布

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本文介绍了一个通过三分法来寻找特定多项式在指定区间内的最小值的问题。该问题涉及数学函数F(x) = 6*x^7 + 8*x^6 + 7*x^3 + 5*x^2 - y*x (0<=x<=100)，并提供了一段C++代码实现。输入为一系列Y值，输出为每组Y对应的函数最小值。

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Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8768 Accepted Submission(s): 5997

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

Author

Redow

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

关于三分求解答案的过程：

1.首先确定图形。

2.确定是求最大值还是最小值。

3.当lm处的值小于rm处的值，则当前最小值在[l,rm]区间，否则在[lm.r]区间；则当前最大值恰好与前面相反。

#include <bits/stdc++.h>
using namespace std;

const double eps = 1e-7;
int t,y;

double val(double x){
    return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
}

int main(){
    scanf("%d",&t);
    while(t --){
        scanf("%d",&y);
        double l = 0, r = 100;
        while(l + eps < r){
            double lm = l + (r - l) / 3;
            double rm = r - (r - l) / 3;
            if(val(lm) < val(rm)) r = rm;
            else l = lm;
        }
        printf("%.4lf\n",val(l));
    }
    return 0;
}