Codeforces 621D Rat Kwesh and Cheese【Long Double】

D. Rat Kwesh and Cheese

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Wet Shark asked Rat Kwesh to generate three positive real numbers
x, y and
z, from 0.1 to
200.0, inclusive. Wet Krash wants to impress Wet Shark, so all generated numbers will have
exactly one digit after the decimal point.

Wet Shark knows Rat Kwesh will want a lot of cheese. So he will give the Rat an opportunity to earn a lot of cheese. He will hand the three numbers
x, y and
z to Rat Kwesh, and Rat Kwesh will pick one of the these twelve options:

a1 = xyz;

a2 = xzy;

a3 = (xy)z;

a4 = (xz)y;

a5 = yxz;

a6 = yzx;

a7 = (yx)z;

a8 = (yz)x;

a9 = zxy;

a10 = zyx;

a11 = (zx)y;

a12 = (zy)x.

Let m be the maximum of all the
ai, and
c be the smallest index (from 1 to
12) such that ac = m. Rat's goal is to find that
c, and he asks you to help him. Rat Kwesh wants to see how much cheese he gets, so he you will have to print the expression corresponding to that
ac.

Input
The only line of the input contains three space-separated real numbers
x, y and
z (0.1 ≤ x, y, z ≤ 200.0). Each of
x, y and
z is given with exactly one digit after the decimal point.

Output
Find the maximum value of expression among xyz,
xzy,
(xy)z,
(xz)y,
yxz,
yzx,
(yx)z,
(yz)x,
zxy,
zyx,
(zx)y,
(zy)x and print the corresponding expression. If there are many maximums, print the one that comes first in the list.

xyz should be outputted as
x^y^z (without brackets), and
(xy)z should be outputted as
(x^y)^z (quotes for clarity).

Examples

Input

1.1 3.4 2.5

Output

z^y^x

Input

2.0 2.0 2.0

Output

x^y^z

Input

1.9 1.8 1.7

Output

(x^y)^z

题目大意：

给你12个数，输出最大的那个。

思路：

浮点数范围：



那么对于200^200的数据是可以用long double来存下来的。

所以我们对12个数都取对数，然后再比较大小即可。

如果取两次对数的话，精度损失会很多。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
long double x,y,z;
long double num[50];
char s[15][15] = {" ","x^y^z", "x^z^y", "(x^y)^z", "(x^z)^y", "y^x^z", "y^z^x", "(y^x)^z", "(y^z)^x", "z^x^y", "z^y^x", "(z^x)^y", "(z^y)^x"
};
int main()
{
while(cin>>x>>y>>z)
{
long double maxn=-1;
char ans[15];
num[1]=pow(y,z)*log(x);
num[2]=pow(z,y)*log(x);
num[3]=y*z*log(x);
num[4]=y*z*log(x);
num[5]=pow(x,z)*log(y);
num[6]=pow(z,x)*log(y);
num[7]=x*z*log(y);
num[8]=x*z*log(y);
num[9]=pow(x,y)*log(z);
num[10]=pow(y,x)*log(z);
num[11]=x*y*log(z);
num[12]=x*y*log(z);
for(int i=1;i<=12;i++)
{
if(num[i]>maxn)
{
maxn=num[i];
strcpy(ans,s[i]);
}
}
cout<<ans<<endl;
}
}