【二分查找】 Can you solve this equation? HDU基础04二分法02_now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x-优快云博客

本文介绍了一种通过迭代逼近的方法来寻找特定区间内多项式方程的实数解的算法实现。该算法适用于形如8*x^4+7*x^3+2*x^2+3*x+6=Y的四次多项式方程，并能在0到100的范围内找到方程的根或报告无解情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

2
100
-4

Sample Output

1.6152
No solution!

Author

Redow

#include <bits/stdc++.h>
using namespace std;

double f(double x)
{
	return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
}
int main()
{
	int T;scanf("%d",&T);
	while(T--)
	{
		long long y;
		scanf("%lld",&y);

		double l=0,r=100;
		double ans=-1;

		for(int i=0;i<100;i++)
		{   //循环一定次数保证精度
			double mid=(l+r)/2;
			if(abs(f(mid)-y)<=1e-6)
			{
				ans=mid;
				break;
			}
			else if(f(mid)>y)
				r=mid;
			else l=mid;
		}

		if(ans==-1)
			printf("No solution!\n");
		else
			printf("%.4f\n",ans);
	}
	return 0;
}