Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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对这个函数连续求几次导可以发现，对于此函数的导函数，如果y <= 0,那么函数在[0, 100]内单调递增，所以最小值是0

如果y>0,观察其导函数，发现导函数单调递增，所以我们只要找到导函数的零点（即原函数的极小值点），此时可以用二分解决

找到极值点以后，如果它在[0, 100]，就把它代入函数，否则说明函数在[0, 100]内单调递减，把100代入函数即可

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
	int t;
	scanf("%d\n", &t);
	while (t--)
	{
		double y;
		scanf("%lf", &y);
		if (y <= 0)
		{
			printf("0.0000\n");
			continue;
		}
		double l = 0, r = 100, mid;
		while(r - l > 1e-6)
		{
			mid = (l + r) / 2;
			double dx = 42.0 * pow(mid, 6) + 48.0 *  pow(mid, 5) + 21.0 *  pow(mid, 2) + 10.0 * mid;
			if (abs(dx - y) < 1e-6)
			{
				break;
			}
			else if(dx - y > 1e-6)
			{
				r = mid;
			}
			else
			{
				l = mid;
			}
		}
		if (mid - 100.0 > 1e-6)
		{
			mid = 100.0;
		}
		double ans = 6 *  pow(mid, 7) + 8 *  pow(mid, 6) + 7 * pow(mid, 3) + 5 *  pow(mid, 2) - y * mid;
		printf("%.4f\n", ans);
	}
	return 0;
}