Strange fuction



Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)


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Total Submission(s): 5095    Accepted Submission(s): 3633




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


 



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解析：设函数 df(x) 为 f(x) 的导数，则有：存在唯一的x0，使得df(x0)==0。

           若0<=x0<=100，ans=min(f(0)，f(100)，f(x0))；否则，ans=min(f(0)，f(100))。

代码：


#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

double precision=1e-6;
double y;

double f(double x)
{
  return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}

double df(double x)
{
  return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y; 
} 

int main()
{
  freopen("1.in","r",stdin);
  
  int t;double l,r,mid,ans;
  scanf("%d",&t);
  while(t--)
    {
      scanf("%lf",&y);
      ans=min(f(0),f(100));
      
      if(df(0)*df(100)>=0)goto d1;
      l=0,r=100;
      while(r-l>=precision)
        {
          mid=(l+r)/2;
          if(df(mid)<=0)l=mid;
          else r=mid;
    }
    ans=min(ans,f(l));
      d1:printf("%.4lf\n",ans);
  }
  return 0;
}