## Intuition The problem requires us to compare two expressions, \(x^y\) and \(y^x\), without directly computing these potentially enormous numbers. Direct computation is infeasible due to the large possible values of \(x\) and \(y\) (up to \(10^9\)). Instead, we can leverage logarithmic properties to transform the comparison into a more manageable form. ## Approach To compare \(x^y\) and \(y^x\), we can use logarithms to avoid dealing with large numbers: 1. Take the natural logarithm of both sides: - \(\ln(x^y) = y \cdot \ln(x)\) - \(\ln(y^x) = x \cdot \ln(y)\) 2. The problem of comparing \(x^y\) and \(y^x\) reduces to comparing \(y \cdot \ln(x)\) and \(x \cdot \ln(y)\). 3. Calculate these two expressions using the `log` function from the C++ standard library, which computes the natural logarithm. 4. Compare the two results: - If \(y \cdot \ln(x) < x \cdot \ln(y)\), then \(x^y < y^x\). - If \(y \cdot \ln(x) > x \cdot \ln(y)\), then \(x^y > y^x\). - If they are equal, then \(x^y = y^x\). This approach avoids overflow and allows us to make the comparison efficiently. ## Complexity - **Time:** \(O(1)\) - The operations involved (logarithm and multiplication) are constant time. - **Space:** \(O(1)\) - We use a fixed amount of extra space for calculations. ## Solution (C++) Here's the complete C++ solution implementing the above approach: ```cpp #include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); long long x, y; cin >> x >> y; if (x == y) { cout << "=" << endl; } else { double lhs = y * log(x); double rhs = x * log(y); if (lhs < rhs) { cout << "<" << endl; } else if (lhs > rhs) { cout << ">" << endl; } else { cout << "=" << endl; } } return 0; } ``` This solution efficiently compares \(x^y\) and \(y^x\) by transforming the problem using logarithms, thus avoiding the need to compute large powers directly. The use of `log` ensures that we can handle the large range of input values without overflow.

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