HDU-2199-Can you solve this equation?【二分】-优快云博客

本文介绍了一种通过二分查找法求解特定多项式方程的数值解的方法，并提供了两段不同的C++代码实现，旨在解决形式为8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = Y的方程，当Y值位于一定范围内时，在0到100之间找到方程的一个实数解。

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Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 16553 Accepted Submission(s): 7343

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!

代码一：

#include<cstdio>
#include<algorithm>
#include<cmath>
#define eps 1e-12
using namespace std;
double f(double x)
{
	return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2)+3*x+6;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		double y,mid,maxn;
		double l=0.0,r=100.0;
		scanf("%lf",&y);
		maxn=8*pow(100.0,4.0)+7*pow(100.0,3.0)+2*pow(100.0,2)+3*100+6;
		if(y<6||y>maxn)
		{
			printf("No solution!\n");
			continue;	
		}
		int size=50;
		while(size--)  //循环次数一定比 50 少 
		{
			mid=(l+r)/2.0;
			if(y<f(mid))	r=mid-eps;  //不加 eps 这个精度也能 AC 
			else	l=mid+eps;
		}
		printf("%.4lf\n",l);
	}
	return 0;
}

代码二：

#include<cstdio>
#include<cmath>
double f(double x)
{
	return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
}
int main()
{
	int t,a;
	scanf("%d",&t);
	while(t--)
	{
		double y,l,r,mid;
		scanf("%lf",&y);
		if(y < 6 || y > 8*100*100*100*100+7*100*100*100+2*100*100+3*100+6)
			printf("No solution!\n");
		else
		{
			l=0;
			r=100;
			a=50;
			mid=(l+r)/2;
			while(fabs(f(mid)-y)>1e-5)
			{
				if(f(mid) < y)
				{
					l=mid+1;
				}
				else
				{
					r=mid-1;
				}
				mid=(l+r)/2;
			}
			printf("%.4f\n",mid);
		}
		
	}
}