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

   
   
   
   

    
    
    
    2
100
200
   
   
   
   

 

 

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; else f=mid; } printf("%.4lf/n",find((f+l)/2,y)); } return 0; }