L1-067 Roche limit (10 points)

L1-067 Roche limit (10 branch )

Science fiction film 《 Wandering the earth 》 One of the most important scenarios in is when the earth is too close to Jupiter , The atmosphere began to be sucked away by Jupiter , And as we get closer to the ground wood “ Rigid body Roche limit ”, The earth is in danger of being completely torn apart . But actually , This calculation is wrong .

Roche limit （Roche limit） It is the distance when the gravity of one celestial body is equal to the tidal force caused by the second celestial body . When the distance between two celestial bodies is less than the Roche limit , Objects tend to disintegrate , And then it becomes the ring of the second object . It was Edward, the first person to calculate this limit · Roche named .（ From baidu baike ）

The ratio of the density of large objects to that of small objects 3 After power , Multiplied by the radius of the big object and a multiple （ The multiple corresponding to the fluid is 2.455, The multiple of a rigid body is 1.26）, Is the value of the Roche limit . For example, the density ratio of Jupiter to earth is open 3 The power is 0.622, If we assume that the earth is a fluid , So the Roche limit is 0.622×2.455=1.52701 Times the radius of Jupiter ; But the earth is a rigid body , The corresponding Roche limit is 0.622×1.26=0.78372 Times the radius of Jupiter , This distance is smaller than the radius of Jupiter , That is, the earth will be torn apart only when it is inside Jupiter , In other words , Is that the earth can't be torn apart .

This question asks you to judge whether a small celestial body will be torn apart by a large celestial body .

Input format ：
Enter... On one line 3 A digital , In turn ： The ratio of the density of large objects to that of small objects 3 The value calculated after the power （≤1）、 Properties of small celestial bodies （0 Represents a fluid 、1 Represents a rigid body ）、 The ratio of the distance between two celestial bodies to the radius of large celestial bodies （>1 But not more than 10）.

Output format ：
In one line, first output the ratio of the Roche limit of small objects to the radius of large objects （ Output after decimal point 2 position ）; Then empty one space ; The final output ^_^ If the little celestial body won't be torn apart , Otherwise output T_T.

sample input 1：

0.622 0 1.4

sample output 1：

1.53 T_T

sample input 2：

0.622 1 1.4

sample output 2：

0.78 ^_^

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<sstream>

using namespace std;
typedef long long ll;
const int N=10010;

int main()
{
    
	double a, b, c;
	cin >> a >> b >> c;
	if(b == 0) a *= 2.455;
	else a *= 1.26;
	if(a < c) printf("%.2lf ^_^\n",a);
	else printf("%.2lf T_T\n",a);
	return 0;
}