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.

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);

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.

2
100
-4

1.6152
No solution!

算法：

	二分法

思路：

	设top,bot,mid三个变量，并且写一个 f(x)的函数来代替上述方程，简化程序

	先判断	f(top)==0? f(bot)==0?

	若都不为零，则进行二分

	每次判断f(top)f(mid)>0? 	f(bot)f(mid)>0?

	若前者小于零，则零点必然在[mid,top]中，则bot=mid，然后继续上述步骤，直到达到题目要求

	若后者小于零，则零点必然在[bot,mid]中，则top=mid，然后继续上述步骤，直到达到题目要求

	否则就不存在零点

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
double f(double x)
{
    double t;
    t=8*pow(x,4) + 7*pow(x,3) + 2*x*x + 3*x + 6;
    return t;
}
using namespace std;
int main ()
{
    double x,y,top=100,bot=0,mid=50,ft,fb,fm;
    int t;
    bool find;
    cin>>t;
    while (t--)
    {
        find=true;
        cin>>y;
        top=100;
        bot=0;
        mid=50;
        ft=f(top)-y;
        fb=f(bot)-y;
        fm=f(mid)-y;
        if (ft==0)
            printf("%.4lf\n",top);
        else if (fb==0)
            printf("%.4lf\n",bot);
        else
        {
            while(fabs(fm)>1e-4)
            {
                ft=f(top)-y;
                fb=f(bot)-y;
                fm=f(mid)-y;
                if (fb*fm<0)
                {
                    top=mid;
                    mid=(top+bot)/2;
                }
                else
                {
                    bot=mid;
                    mid=(top+bot)/2;
                }
                if (fabs(mid-100)<1e-4)
                {
                    find=false;
                    break;
                }


            }
            if (find==true)
                printf("%.4lf\n",mid);
            else
                cout<<"No solution!"<<endl;
        }
    }
    return 0;
}