Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26314 Accepted Submission(s): 11274
Problem Description
Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 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!
Author
Redow

題目的大意是 給定一個關於x的函式f(x)，然後再給定一個Y，讓你在 0<=x <=100的範圍內找出一個x使得函式f(x)等於Y
很容易看出來這個函式在0,100間是單調遞增的（那麼我們就採用二分法）

對x的區間不斷的進行二分，直到fab(f(x)-Y)的值逼近一個EPS（一個很小很小的數字）就可以了

AC程式碼如下：

#include <iostream>
#include <cmath>
#include <iomanip>
 

using namespace std;
double Y;
double Function(double x)
{
    return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;  //用來計算函式值
}
const double EPS=1e-5; //控制精度
double twofen(double low,double high) //二分
{
    double mid=(low+high)/2; 
    while(fabs(Function(mid)-Y)>EPS)
    {
        if(Function(mid)>Y) //因為是單調遞增的函式，所以如果f(mid)的值比Y大，說明mid在我們想找的x的右邊
            high=mid;
        else low=mid; //反之
        mid=(low+high)/2;
    }
    return mid;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>Y;
        cout<<fixed<<setprecision(4);  //控制資料精度為小數點4位
        if(Function(0)>Y || Function(100)<Y) //因為是單調遞增的函式，所以如果 當 x=0大於Y的時候，和X=100 小於Y的時候不可能找到解
            cout<<"No solution!"<<endl;
        else cout<<twofen(0,100)<<endl;
    }
    return 0;
}