acm_专题2_1001

题目描述：

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;<br>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<br>100<br>-4<br>
  
  

 

Sample Output

  
  

   
   1.6152<br>No solution!<br>
  
  

 

Author

Redow

 

题目大意：

就是给你一个函数，把函数值给你，让你求出自变量的值。。

想法：

这个题函数是递增的，在（0，100）内      所以可以用二分查找    另精度问题可用<iomanip>头文件解决。。

代码：

#include <iostream>

#include<iomanip>
#include <cmath>
using namespace std;
double v(double x)
{
    return (8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6);
}
double f(double x,double z, double y)
{
    double mid;
    while((y-z)>1e-10)
    {
        mid=(z+y)/2;
        if(v(mid)<x)
            z=mid+1e-10;
        else
            y=mid-1e-10;
    }
    return mid;
}
int main()
{
    int n;
    double a;
    cin>>n;
    while(n--)
    {
        cin>>a;
        if(a<6||a>807020306||fabs(f(a,0,100))<1e-4)
        cout<<"No solution!"<<endl;
        else
        cout<<setprecision(4)<<fixed<<f(a,0,100)<<endl;
    }

}

感想：

第一次用二分查找，感觉挺好用的。。