HDU 2199 Can You solve this equation? Two points

The main is to practice the next two points of thought + floating point precision control

Can You solve this equation? Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 7029 Accepted Submission (s): 3266


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 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);
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!
The monotonicity of this equation is very good proof, the derivation is easy to obtain, in the monotone solution with two points, the main is to try a variety of precision control.

#include <cstdio>
#include <cmath>
#include <cstdlib>

const double EPS = 1e-7;

Double cal (double x) {
	return 8*pow (x,4.0) +7*pow (x,3.0) +2*pow (x,2.0) +3*x+6.0;
}

int main () {
	int t;
	Double low,high,mid,y,res;

	scanf ("%d", &t);
	while (t--) {
		scanf ("%lf", &y);
		if (y<cal (0.0) | | y>cal (100.0)) {
			printf ("No solution!\n");
			Continue;
		}
		low = 0.0;
		High = 100.0;
		while (high-low>eps) {
			mid = (Low+high)/2;
			res = cal (mid);
			if (res<y) {Low
				= mid + 1e-7;
			} else{High
				= mid-1e-7;
			}
		}
		printf ("%0.4lf\n", mid);
	}
	return 0;
}