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B. High School: Become Humantime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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?$x^y$

?with? $y^x$

. 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?$x^y$

?with? $y^x$

?for Vasya, maybe then other androids will respect him.

Input

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

?and? $y$?( $1\le x,y\le {10}^9$

).

Output

If?$x^y<y^x$

, then print ' <' (without quotes). If? $x^y>y^x$, then print ' >' (without quotes). If? $x^y=y^x$

, then print '=' (without quotes).

ExamplesinputCopy5 8 outputCopy> inputCopy10 3 outputCopy< inputCopy6 6 outputCopy= Note

In the first example?$5^8=5?5?5?5?5?5?5?5=390625$

, and? $8^5=8?8?8?8?8=32768$

. So you should print '>'.

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

.

In the third example?$6^6=46656=6^6$

.

題意：

輸入兩個數，x,y，如果x^y>y^x就輸出>，小于就輸出<，等于就輸出=。

思路：

由數據范圍，肯定不能乘算，所以可以變換一下? ?兩邊取對數 log（x^y） 與 log（y^x）比較

即：y*log（x） 與 x*log（y）比較

#include <bits/stdc++.h> using namespace std; #define eps 1e-6 int main() {int x,y;double xx;scanf("%d %d",&x,&y);xx = y*log10(x) - x*log10(y);if(fabs(xx) <= eps)printf("=\n");else{if(xx < eps)printf("<\n");elseprintf(">\n");}return 0; }

大佬的AC代碼

#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("="); }

錯誤

若x=2,y=4，則2^4=4^2

#include <bits/stdc++.h> using namespace std; #define eps 1e-6 int main() {int x,y;double xx;scanf("%d %d",&x,&y);if(x==y)printf("=\n");else{xx = y*log10(x) - x*log10(y);if(xx < eps)printf("<\n");elseprintf(">\n");}return 0; }

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