Can you solve this equation?

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!

题意：利用二分法求解；

解题思路：用最简单的二分就能过，但是！！！注意精度最少1e-6；

感悟：终于知道学长说的，二分最蛋疼的就是控制精度了；

代码：

#include

#include

using namespace std;

double iteration(double x)

{  

    double
fx;

   
fx=8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;

    return
fx;

}

int main()

{

   
//freopen("in.txt", "r", stdin);

    int n;

    double
y,x1,x2,x;

   
scanf("%d",&n);

    for(int
i=0;i

    {

       
scanf("%lf",&y);

       
x1=-1;

       
x2=101;

       
if(y<6||y>807020306)

       
{

           
printf("No solution!\n");

           
continue;

       
}//最大解和最小解判断有没有解

       
else

       
{

           
while(true)

           
{

               
x=(x1+x2)/2;

               
if(x2-x1<1e-6)//这里精度最小就得是1e-6

               
{

                   
printf("%.4lf\n",x);

                   
break;

               
}

               
else

               
{

                   
if(iteration(x)==y)

                   
{

                       
printf("%.4lf\n",x);

                       
break;

                   
}

                   
else if(iteration(x)

                       
x1=x;

                   
else if(iteration(x)>y)

                       
x2=x;

              
}

           
}

       
}

    
}

    
return 0;

}