HDU 2899 Strange fuction

最新推荐文章于 2022-07-19 16:47:04 发布

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本博客探讨如何通过输入实数y，求解函数F(x)=6*x^7 + 8*x^6 + 7*x^3 + 5*x^2 - y*x在区间[0,100]内的最小值，提供了二分法和三分法两种解题策略。

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A - Strange fuction

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2899

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

Sample Output

二分三分都可以做，三分更快些。

二分的做法是求导函数。然后求导函数为0的点的值便是最小值。

1.二分法（15ms）

#include<stdio.h>
#include <math.h>
double F1(double x, double y)
{
	return 42 * pow(x, 6) + 48 * pow(x, 5) + 21 * pow(x, 2) + 10 * x - y;
}
double calculate(double x, double y)
{
	return 6 * pow(x, 7) + 8 * pow(x, 6) + 7 * pow(x, 3) + 5 * pow(x, 2) - y*x;
}
int main()
{
	int ynum;
	scanf("%d", &ynum);
	while (ynum--)
	{
		double Y;
		scanf("%lf", &Y);
		double lit = pow(10.0, -5);
		double min = 0, max = 100;
		double mid = (max + min) / 2;
		while (min < max&&fabs(F1(mid, Y)) > lit)
		{
			if (F1(mid, Y) > 0)
			{
				max = mid;
			}
			else if (F1(mid, Y) < 0)
			{
				min = mid;
			}
			else
				break;
			mid = (max + min) / 2;
		}
		if (mid>0 && mid<100)
		{
			printf("%.4f\n", calculate(mid,Y));
		}
		else
		{
			printf("0.0");
		}
	}
	return 0;
}

2.三分法（0ms）

#include <stdio.h>
#include <math.h>
#define eps 1e-8
double y;
double cal(double x)
{
        return  6 * pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
int main()
{
        int t;
        scanf("%d",&t);
        while(t--)
        {
                scanf("%lf",&y);
                double low=0,high=100;
                double mid, midmid;
                while ( low + eps < high )
                {
                    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(high));
        }
        return 0;
}