987B - High School: Become Human

B. High School: Become Human
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.

It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.

One of the popular pranks on Vasya is to force him to compare xyxy with yxyx. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.

Please help Vasya! Write a fast program to compare xyxy with yxyx for Vasya, maybe then other androids will respect him.

Input

On the only line of input there are two integers xx and yy (1x,y1091≤x,y≤109).

Output

If xy<yxxy<yx, then print '<' (without quotes). If xy>yxxy>yx, then print '>' (without quotes). If xy=yxxy=yx, then print '=' (without quotes).

Examples
input
Copy
5 8
output
Copy
>
input
Copy
10 3
output
Copy
<
input
Copy
6 6
output
Copy
=
Note

In the first example 58=55555555=39062558=5⋅5⋅5⋅5⋅5⋅5⋅5⋅5=390625, and 85=88888=3276885=8⋅8⋅8⋅8⋅8=32768. So you should print '>'.

In the second example 103=1000<310=59049103=1000<310=59049.

In the third example 66=46656=6666=46656=66.


题意:输入两个数,x,y,如果x^y>y^x就输出>,小于就输出<,等于就输出=。

题解:数论 取对数,取 的对数有精度误差,做计算的时候判断注意除和减,不然很容易错。直接比较就不会了。

我最开始想的是找规律,有一种2 4就难受了,然后最近学python,python里面不是大数据无所畏惧么,暴力了一发,然而RE。听大佬提示,还是老老实实用对数吧。

AC代码c++:根据留言情况,可能在一些编译器上第三个样例 6 6在一些编译器上输出的结果不对,应该是编译器的问题,例如CB,但是理论和CF都能AC,我还再次提交一次,AC了。ps:2018年6月2 日

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    if(log(x)*y>(log(y)*x))puts(">");
    else if(log(y)*x>(log(x)*y)) puts("<");
    else puts("=");
}
import math
x,y=map(int,input().split(' '))
if y*math.log(x)>x*math.log(y):
        print(">")
elif y*math.log(x)<x*math.log(y):
          print("<")
else:
        print("=")


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转载自blog.csdn.net/memory_qianxiao/article/details/80504837