HDU 2675 Equation Again 二分求函数零点

/*
取对数后得
ln(x)/x=(1+ln(y))/(e*y)
显然右边是常数
作图可知道ln(x)/x在x>=1范围先增后减，极值在x=e处
对于任意一个y!=1，都有左右两个解，二分解决就是了……
*/
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
const double ee=2.718281828459;
const double eps=1e-7;
int main()
{
    double x,y;
    // for(int i=1;i<=2000;i++)
    // cout<<log(i)/(1.0*i)<<endl;
    // while(scanf("%lf",&y)!=EOF)
    while(scanf("%lf",&y)!=EOF)
    {
        double tmp=(1+log(y))/(y*ee);
        double low=1+eps,hei=ee-eps,mid;
        if(tmp*ee-1>eps){puts("Happy to Women’s day!");continue;}
        while(hei-low>eps)
        {
            mid=(hei+low)/2;
            if(log(mid)*(y*ee)>(1+log(y))*mid)
            hei=mid;
            else 
            low=mid;
        }
        if(y==1)printf("%.5lf\n",(hei+low)/2);
        else  if(y>1)
        {
            double ans=(hei+low)/2;
            low=ee+eps,hei=2000000000+eps;
            while(hei-low>eps)
            {
                mid=(hei+low)/2;
                if(log(mid)*(y*ee)<(1+log(y))*mid)
                hei=mid;
                else 
                low=mid;
            }
            printf("%.5lf %.5lf\n",ans,(hei+low)/2);
        }
    }
    return 0;
}