HDOJ 2899 Strange fuction（二分，求导。。）

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4287    Accepted Submission(s): 3092

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

 

 

求函数x在区间[0,100]内的最小值，可以把y看成一个实数。用到了求导，二次求导。表示数学很渣，高数课也没怎么去，高中学的也全忘光了。不太懂啊。。。。。

 

求导结束，判断单调性。是一个凹函数，可以用二分法求极值。

 

具体代码如下：

 

#include<stdio.h>#include<math.h>#define res 1e-8double y;double g(double x){return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x;//一次求导 }double f(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;double l,r,m;scanf("%d",&t);while(t--){scanf("%lf",&y);if(g(100.0)-y<=0)/*g(x)为递增的，-y后如果为0，则整个f(x)为递减*/  {printf("%.4lf\n",f(100.0));continue;} /*算二次导数，可知f(x)为凹函数，存在最小值，然后二分求解*/  l=0;r=100;while(r-l>=res){m=(r+l)/2;if(g(m)-y<0)   l=m;else   r=m;}printf("%.4lf\n",f(m));}return 0;}