[hdu]can you solve this equation

最新推荐文章于 2025-12-24 18:35:39 发布

原创最新推荐文章于 2025-12-24 18:35:39 发布 · 307 阅读

0 ·

CC 4.0 BY-SA版权

二分法专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个特定方程的数值求解方法，使用二分查找算法寻找方程在0到100范围内的解，并给出了完整的C++实现代码。

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11569 Accepted Submission(s): 5305

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!

题意：求x使得满足题目所给的方程。

思路：最简单的二分查找，要注意小数点位数。

#include "cstdio"
#include "cmath"
#define mi 0.00000001
using namespace std;
double f(double x)
{
	return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		double y;
		scanf("%lf",&y);
		if(f(0)>y||f(100)<y)
		{
			printf("No solution!\n");
			continue ;
		}
		double l=0,r=100;
		while(r-l>mi)
		{
			double mid=(r+l)/2;
			if(f(mid)<=y)l=mid;
			else r=mid;
		}
		printf("%.4f\n",(r+l)/2);
	}
	return 0;
}