Hdoj 2899.Strange fuction 题解
Problem Description
Now, here is a fuction:
F(x) = 6 * x7+8*x6+7x3+5*x2-yx (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
思路
函数是：\(F(x) = 6x^7+8x^6+7x^3+5x^2-yx\)
导函数是：\(F\'(x) = 42x^6 +48x^5+21x^2+10x-y\)
由题目条件可得，导函数单调递增，\(F\'(100)\)最大，如果它还小于0，说明原函数单调递减，\(F(100)\)最小。其他情况是：二分查找导函数的零点，找到后代入原函数就可以得到答案
代码
#include<bits/stdc++.h>
using namespace std;
double y;
double f(double x)
{
return 6*pow(x,7) + 8*pow(x,6) + 7*pow(x,3) + 5*pow(x,2) - y*x;
}//函数
double df(double x)
{
return 42*pow(x,6) + 48*pow(x,5) + 21*pow(x,2) + 10*x - y;
}//导函数
int main()
{
int n;
cin >> n;
while(n--)
{
cin >> y;
if(df(100) <= 0)
{
printf("%.4lf\n",f(100));
continue;
}
double l = 0, r = 100.0;
double mid;
while(r-l>=1e-8)
{
mid = (l+r)/2;
if(df(mid)<0)
l = mid;
else
r = mid;
}
printf("%.4lf\n",f(mid));
}
return 0;
}
版权声明：本文为MartinLwx原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。