high school d×d第二季啥时候出

Posted 2023-04-15

参考技术A 官方已经确定会出第2季，希望制作阵容不变，故事内容紧接第一季！播出具体时间未定！

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High School: Become Human（数学思维）

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 ${\mathrm{xyxy with yxyx.\; Other\; androids\; can\; do\; it\; in\; milliseconds\; while\; Vasya‘s\; memory\; is\; too\; small\; to\; store\; such\; big\; numbers.}}^{}$

Please help Vasya! Write a fast program to compare ${\mathrm{xyxy with yxyx for\; Vasya,\; maybe\; then\; other\; androids\; will\; respect\; him.}}^{}$

Input

On the only line of input there are two integers $\mathrm{xx and yy (1\le x,y\le 1091\le x,y\le 109).}$

Output

If ${\mathrm{xy<yxxy<yx,\; then\; print\; ‘<‘\; (without\; quotes).\; If xy>yxxy>yx,\; then\; print\; ‘>‘\; (without\; quotes).\; If xy=yxxy=yx,\; then\; print\; ‘=‘\; (without\; quotes).}}^{}$

Examples

input

Copy

5 8

output

Copy

>

input

Copy

10 3

output

Copy

<

input

Copy

6 6

output

Copy

=

Note

In the first example ${\mathrm{58=5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=39062558=5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=390625,\; and 85=8\cdot 8\cdot 8\cdot 8\cdot 8=3276885=8\cdot 8\cdot 8\cdot 8\cdot 8=32768.\; So\; you\; should\; print\; ‘>‘.}}^{}$

In the second example ${\mathrm{103=1000<310=59049103=1000<310=59049.}}^{}$

In the third example ${\mathrm{66=46656=6666=46656=66.}}^{}$

${\mathrm{题目意思：给你两个数x,\; y,\; 比较\; x^y\; 和\; y\; ^\; x\; 的大小}}^{}$

${\mathrm{解题思路：}}^{}$

 

#include<cstdio>
#include<cmath>
int main()
{
    int x,y;
    scanf("%d%d",&x,&y);
    if(log(x)*y>log(y)*x)
    {
        printf(">
");
    }
    else if(log(x)*y<log(y)*x)
    {
        printf("<
");
    }
    else
    {
        printf("=
");
    }
   return 0;
}