This problem’s author is too lazy to write the problem description, so he only give you a equation like X (eY) == (eY) x, and the value of Y, your task is calculate the value of X.
Note : here e is the Natural logarithm.
Input
Each line will contain one number Y(Y >= 1). Process to end of file.
Output
For each case, output X on one line, accurate to five decimal places, if there are many answers, output them in increasing order, if there is no answer, just output “Happy to Women’s day!”.
Sample Input
1
Sample Output
2.71828
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define ll long long int
#define ull unsigned long long int
#define e 2.718281828459
#define INF 0x7fffffff
#pragma warning(disable:4996)
#define pf printf
#define sf scanf
#define min(a,b) (a)>(b)?(a):(b);
double PI = acos(-1.0);
double eps=1e-10;
bool Judge(double x, double y) {
if (x / log(x) >= e * y / log(e * y))
return 1;
return 0;
}
int main(void) {
double ans1=0.0, ans2=0.0;
double y;
while (sf("%lf",&y)!=EOF) {
double left = 1.0, right = e;
double mid;
if (y / exp(y) <= 1.0 / e) {
while (right-left>eps) {
mid = left + (right-left) / 2;
if (Judge(mid, y)) {
left = mid;
}
else {
right = mid;
}
}
ans1 = left;
left = e, right = INF;
while (right- left > eps) {
mid = left+(right - left) / 2;
if (Judge(mid,y)) {
right = mid;//函数增减性转变
}
else {
left = mid;//函数增减性转变
}
}
ans2 = left;
if (ans2- ans1>1e-5)
pf("%.5lf %.5lf\n", ans1, ans2);
else
pf("%.5lf\n", ans1);
}
else
pf("Happy to Women’s day!\n");
}
return 0;
}