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#题目：

Now, here is a fuction:

F(x) = 6 * x ^ 7+8x^ 6+ 7x ^ 3+5x^2-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

2100200

Sample Output

-74.4291-178.8534

解题思路：这个题就是给一个函数式，然后由x在区间0-100之间取值求最小值。

利用函数求导为：42*x^6+48 *x ^5+21 *x^2+10 *x-y==0,利用二分求x的值，使倒数为0，则函数值最小。

程序代码：

#include
   
    #include
    
     double l=0.00000001;int main(){
   	double x,m,n,k,y,sum;	int T;	scanf("%d",&T);	while(T--)	{
   		scanf("%lf",&y);		n=0.00;		m=100.00;//x的区间是0-100		while(m-n>l)//让n和m无限接近，使得导函数为0		{
   			k=(m+n)/2;			if(42*pow(k,6)+48*pow(k,5)+21*pow(k,2)+10*k-y>0)				m=k;			else				n=k;		}		sum=6*pow(k,7)+8*pow(k,6)+7*pow(k,3)+5*pow(k,2)-y*k;		printf("%.4f\n",sum);	}	return 0;}
    
   

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