hdu2199

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!
  
  

 

本来是按直接暴力打的，但是一直WA。后来百度了一下发现二分居然可以按照精度查！

#include<iostream>
#include<stdlib.h>
#include<string>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
double sen(double m){
    return    8*m*m*m*m + 7*m*m*m + 2*m*m + 3*m + 6; 
}
double bs(double x){
    double l=0,r=100,mid=(l+r)/2;
    while(fabs(sen(mid)-x)>1e-5){
        mid=(l+r)/2;
        if(sen(mid)>x)r=mid-1e-5;
        else l=mid+1e-5;
    }
    return mid;
}
int main(){
#ifdef LOCAL
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif
    int T;
    cin>>T;
    while(T--){
        double x;
        cin>>x;
        if(x<sen(0)||x>sen(100))printf("No solution!\n");
        else {
            double ans=bs(x);
            printf("%.4lf\n",ans);
        }
    }
    return 0;
}