YT14-HDU-在0-100间找到那个正确的x

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!

代码如下：

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
double Y;
double Equ(double x)
{
    return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;             //判断是否等式是否相等
}

double Search()                                               //二分查找
{
    double high=100,low=0;
    double temp;
    while (high-low>1e-6)
    {
        temp=(low+high)/2;
        if (Equ(temp)<Y)
            low=temp+1e-7;
        else
            high=temp-1e-7;
    }
    return (high+low)/2.0;
}

int main()
{
    int T;
    cin>>T;
    while (T--)
    {
        cin>>Y;
        if (Equ(0)<=Y&&Equ(100)>=Y)                    //如果满足条件，说明0到100中间肯定存在一个值使得Equ(x)=Y
            cout<<setiosflags(ios::fixed)<<setprecision(4)<<Search()<<endl;
        else
            cout<<"No solution!"<<endl;
    }
    return 0;
}

解题思路：

题目大意是在0到100间找到满足 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y 的那个x。

简单的二分查找，重点是计算精度和输出格式。