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

Bismillahir Rahmanir Rahim
If x and y are large numbers, you can use logarithms:
Comparing x^y versus y^x means comparing y*log(x) versus x*log(y)
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll x,y;
    cin>>x>>y;
    if((log(x)*y)==(log(y)*x) || x==y) cout<<"="<<endl;
    else if((log(x)*y)<(log(y)*x)) cout<<"<"<<endl;
    else if((log(x)*y)>(log(y)*x)) cout<<">"<<endl;
    return 0;
}
///Alhamdulillah

Problem Link: https://codeforces.com/contest/987/problem/B

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