Codeforces Round #485 (Div. 2)-B-High School: Become Human

时间：2018-05-31 00:41:40 阅读：262 评论：0 收藏：0 [点我收藏+]

标签：div atom ios code exp input only rod sig

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 $x^{y}$

Please help Vasya! Write a fast program to compare $x^{y}$

Input

On the only line of input there are two integers $x$

Output

If $x^{y} < y^{x}$

Examples

input

Copy

5 8

output

Copy

input

Copy

10 3

output

Copy

input

Copy

6 6

output

Copy

Note

In the first example $5^{8} = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$

In the second example $10^{3} = 1000 < 3^{10} = 59049$

In the third example $6^{6} = 46656 = 6^{6}$

不知道为什么。。。取对数就莫名wa。。。比较x*logy和y*logx就是错的。。。而比较x/y和logx/logy就是对的。。。。

没考虑到一个问题就是x^y=y^x有可能x!=y....所以应该先判读大于和小于最后否则就是等于，这样就避免了考虑精度问题由于本题是1e9，因此x/y-logx/logy=exp()<=1e-12

一般精度是要高于范围三位

 #include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
int main(){
    int x,y;
    while (~scanf("%d%d",&x,&y)){
         double a=log(x)/log(y);
         double b=x*1.0/y;
         if(a>b){
             printf(">\n");
         }else if(a<b){
             printf("<\n");
         }
         else printf("=\n");
    }
    return 0;
}

Codeforces Round #485 (Div. 2)-B-High School: Become Human

标签：div atom ios code exp input only rod sig

原文地址：https://www.cnblogs.com/bluefly-hrbust/p/9114003.html

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