hdu:2199 Can You solve this equation? (two points) __ two points

Can You solve this equation? Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 16582 accepted Submission (s): 7365


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 10 0;
Now, please try your lucky.
Input the ' 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, your should just output one real number (accurate up to 4 decimal places), which is the solution O f 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
Recommend

Lcy

The main idea: give you a Y and then let you in [0,100] This interval to find the feasible solution of the Chinese child.

Solution: 0 and 100 respectively into the upper-type, to find no solution area, if the input does not meet, the direct output no solution. And then l=0,r=100 this interval to solve the two points.

The code is as follows:

#include <cstdio>
#include <cmath>
double y;
BOOL Judge (double X)
{
	if (8.0*pow (x,4.0) +7.0*pow (x,3.0) +2.0*pow (x,2.0) +3.0*x+6.0) >y) return
	true;
	else return
	false;
int main ()
{
	int t;
	scanf ("%d", &t);
	while (t--)
	{
		scanf ("%lf", &y);
		if (y<6| | Y> (8.0*pow (100.0,4.0) +7.0*pow (100.0,3.0) +2.0*pow) (100.0,2.0))//Find no +3.0*100+6.0 
		{
			printf ("No solution!\n");
		}
		else
		{
			double l=0,r=100;
			Double mid;
			int size=100;
			while (size--)//100 times definitely can get the accuracy of the problem required 
			{
				mid= (l+r)/2.0;
				if (judge (mid))
				{
					R=mid
				}
				else
				{
					l=mid;
				}
			}
			printf ("%.4lf\n", mid);
		}
	return 0;
}