Can you solve this equation?

最新推荐文章于 2021-08-29 09:01:50 发布

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本文介绍了一种通过三分法和二分法解决特定四次多项式方程的问题，该方程为8*x^4+7*x^3+2*x^2+3*x+6=Y，在区间[0,100]内寻找方程的实数根。

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Problem Description

Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

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 a real number Y (fabs(Y) <= 1e10);

Output

For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input

Sample Output

  
   1.6152
No solution!

已知y要求x的话，肯定是一个x，函数是一个单调的而不是像二次函数那样，一个y有两个x的结果。刚开始没有想到，把他当成二次，用的三分做的，结果过了。过了之后看的小伙伴的方法其实这道题可以直接二分。

#if 0
#include<iostream>
#include<cmath> 
#include<iomanip>
using namespace std;
double x,y;
double fun(double x)
{
	double sum=(8.0*pow(x,4)+7.0*(pow(x,3))+2.0*pow(x,2)+3.0*x+6.0);
	return sum;
}

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		cin>>y;
		
		double mid,midmid,l=0,r=100;
		while(r-l>1.0e-6) 
		{
			mid=(l+r)/2; 
			midmid=(mid+r)/2; 
			if(fabs(fun(mid)-y)<fabs(fun(midmid)-y))  //
			{
				r=midmid;  
			}
			else 
			{
				l=mid; 
			}
		}
		
		int x=r;
		if(x>0&&x<100) 
		{
			cout<<fixed<<setprecision(4)<<r<<endl;
		}
		else
		{
			cout<<"No solution!"<<endl;
		} 
	} 
		
}
#endif