HDU 2199 (二分法)

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const double ep = 1e-6;

double deal(double n)
{
          return 8*pow(n,4) + 7*pow(n,3) + 2*pow(n,2) + 3*n + 6;

}

int main()
{
    int t,i;
    double y;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%lf",&y);
        double lo = 0,hi = 100;
        if(y<deal(0) || y>deal(100))
        {
                              printf("No solution!\n");
                              continue;
        }
        else
        {
                  double mid;
                  while(hi-lo>ep)
                  {
                            mid = (lo+hi)/2;
                            if(deal(mid)>y)
                            {
                                      hi = mid;
                                        }
                                        else
                                        {
                                                  lo = mid;
                                        }
                              }
                              printf("%.4lf\n",hi);
                    }

    }
    return 0;
}