【精度问题】【HDU2899】Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3809    Accepted Submission(s): 2760

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

 

很水的题 但是WA了几次

虽说最后要求1e-4的精度 

但是因为求最小值 所以注意所求的x值精度 至少要1e-6 所以请注意

#include <cstdio>  
#include <cstdlib>  
#include <cmath>  
#include <cstring>  
#include <ctime>  
#include <algorithm>  
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313   
using namespace std;
double y;
void init()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
}
double cal1(double x)
{
	return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double  cal2(double x)
{
	return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
void solve()
{
	double L=0,R=100,ans=1e10; 
	double temp1=cal1(L),temp2=cal1(R);
	if(temp1<ans) ans=temp1;
	if(temp2<ans) ans=temp2;
	double ANS;
	while(R-L>1e-6)
	{
		double m=(R+L)/2;
		if(cal2(m)>0) R=m;
		else if(cal2(m)<0) L=m;
		else { ANS=m;break; }
	}
	ANS=R;
	if(cal1(ANS)<ans) ans=cal1(ANS);
	printf("%.4lf
",ans);

}
int main()
{
//	init();
	int T;
	cin>>T;
	while(T--)
	{
		cin>>y;
		solve();
	}
	return 0;
}