HDOJ 2899 二分算法

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

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 350Accepted Submission(s): 275

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

思路：

这个题是叫我们求最小值，那么很明显对于函数求最值，最简单的方法就是求导

我们队f(x)求导之后得到：f'(x)=42*x^6+48*x^5+21*x*x+10x-y

很明显，当f'(x)=0时取得最值，然后由于定义域x属于[0,100]那么可以推断，f'(x)=0时

f(x)有最小值。

我们可以利用二分算法去计算f'(x)=0时x为多少（注意f'(x)是单调递增的）

计算出x之后再代入原函数解出最小值即可。

我的代码：

#include<stdio.h> #include<math.h> double df(double x) { return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x; } double f(double x,double y) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x; } int main() { int t,i; double y,left,right,mid; scanf("%d",&t); while(t--) { scanf("%lf",&y); left=0,right=100; for(i=1;i<=100;i++) { mid=(left+right)/2; if(df(mid)>y) right=mid; else left=mid; } printf("%.4lf\n",f(mid,y)); } return 0; }