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

Source: Internet
Author: User
Tags pow
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;
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;
}```

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