HDU 2199 Can You solve this equation? (water problem?!) ）

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); Output:for Each test case, you 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:2100-4 Sample Output:1.6152no solution!

Test instructions: Enter the value of Y, the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = y solution, if there is a solution, then the solution must be in [0, 100] between, Output X, if no solution then output no solution!.

#include <stdio.h>#include<math.h>intMain () {DoubleA, B, C, Y, sum, D; intT, X; scanf ("%d", &T);  while(t--) {scanf ("%LF", &y); X=0; A=0; b= -; D=8*b*b*b*b +7*b*b*b +2*b*b +3*b +6; C=8*a*a*a*a +7*a*a*a +2*a*a +3*a +6; if(y >= c && y <=d) x=1; if(x) { while(B-a > 1e-6)///floating-point number to determine whether B is greater than a{C= (a+b)/2; Sum=8*c*c*c*c +7*c*c*c +2*c*c +3*c +6; if(Sum >y) b= c-1e-7; Elsea= c+1e-7; } printf ("%.4lf\n", (A+B)/2); }        if(x = =0) printf ("No solution!\n"); }    return 0;}

HDU 2199 Can You solve this equation? (water problem?!) ）