HDOJ2199Can you solve this equation?（二分）

最新推荐文章于 2018-11-15 09:36:00 发布

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本文介绍了一种解决特定多项式方程8*x^4+7*x^3+2*x^2+3*x+6=Y的方法，当Y的绝对值小于等于1e10时，在0到100之间寻找方程的解。使用二分查找法逼近解，并提供了实现代码。

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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): 14379 Accepted Submission(s): 6421

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

注意精度即可。

#include<stdio.h>
#include<math.h> 
double fx(double x)
{
	return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
} 
int main()
{
	int test;
	double k,j,y;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%lf",&y);
		if(y>=fx(0)&&y<=fx(100))
		{
			k=0;
			j=100;
			while(j-k>=1e-7)
			{
				if(fx((k+j)/2)<y)
					k=(k+j)/2;
				else
				    j=(k+j)/2;
			}
			printf("%.4lf\n",(j+k)/2);
		}
		else
		printf("No solution!\n");
	}
	
}