HDU-Can you solve this equation? -二分法求高次方程近似解

问题及代码：

Problem F Can you solve this equation? 

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 4

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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!

/*    
*Copyright (c)2015,烟台大学计算机与控制工程学院    
*All rights reserved.    
*文件名称：HDU.cpp    
*作    者：单昕昕    
*完成日期：2015年2月27日    
*版 本 号：v1.0        
*/ 
#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
double f(double x)
{
    return (8*x*x*x*x+7*x*x*x+2*x*x+3*x+6);
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        double y,m=0.0,n=100.0;
        cin>>y;
        if(f(0)>y||f(100)<y)
        {
            cout<<"No solution!"<<endl;
            continue;
        }
        double mid=50.0;
        while(fabs(f(mid)-y)>1e-5)
        {
            if((f(mid)>y))
            {
                n=mid;
                mid=(m+n)/2;
            }
            else if(f(mid)<y)
            {
                m=mid;
                mid=(m+n)/2;
            }
        }
        printf("%.4lf\n",mid);
    }
    return 0;
}

运行结果：

知识点总结：
二分法求高次方程近似解。

学习心得：

主要是二分的思想，用一个自定义函数不断逼近方程的解。