HDU - 2199 ：Can you solve this equation? （二分）

最新推荐文章于 2021-03-24 14:23:15 发布

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最新推荐文章于 2021-03-24 14:23:15 发布

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原文链接：http://www.cnblogs.com/xibeiw/p/7239978.html

本文介绍了一种使用二分法求解特定多项式方程近似根的方法，并提供了一个C++实现示例。该算法适用于在0到100区间内寻找方程8*x^4+7*x^3+2*x^2+3*x+6=Y的实数解。

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.InputThe 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);

OutputFor 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

2
100
-4

Sample Output

1.6152
No solution!

根据数学表达式用二分方法求近似解。

源代码：
#include<iostream>
#include<algorithm>
#include <iomanip>

using namespace std;

int main()
{
int T,a;
cin>>T;
while(T--)
	{
cin>>a;
		a=a-6;
double low=0,hign=100,x,z;

if(a<0 || a>807020300)
cout<<"No solution!"<<endl;
else
			{
while((hign-low)>1e-10)
				{
					x=(low+hign)/2;
					z=8*x*x*x*x + 7*x*x*x+ 2*x*x + 3*x;
if(z>a)
						hign=x;
if(z==a)
break;
if(z<a)
						low=x;
				}
cout<< setprecision(4)<<fixed<<x<<endl;
			}

	}

return 0;
}