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 (1≤x,y≤1091≤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=5?5?5?5?5?5?5?5=39062558=5?5?5?5?5?5?5?5=390625, and 85=8?8?8?8?8=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.
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;
}
原文地址：https://www.cnblogs.com/bluefly-hrbust/p/9114003.html