HDU 2899（搜索题，二分）

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 80    Accepted Submission(s): 62

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

 

利用函数的单调性求极值

f(x)'=g(x)-y

#include <iostream>using namespace std;double POW(double x,int n){int i;double sum=1.0;for(i=1;i<=n;i++)sum*=x;return sum;}double find(double x,double y){return 6*POW(x,7)+8*POW(x,6)+7*POW(x,3)+5*POW(x,2)-y*x ;}double find1(double x){return 42*POW(x,6)+48*POW(x,5)+21*POW(x,2)+10*x;}int main(){int t;double y,f,l,mid;scanf("%d",&t);while(t--){scanf("%lf",&y);f=0;l=100;if(find1(f)-y>=0){printf("%.4lf/n",find(f,y));continue;}else if(find1(l)-y<=0){printf("%.4lf/n",find(l,y));continue;}while(l-f>1e-6){mid=(f+l)/2;if(find1(mid)-y>=0)l=mid;elsef=mid;}printf("%.4lf/n",find((f+l)/2,y));}return 0;}