a ^ b或b ^ a中的较大者(a提升为b的幂或b提升为a的幂)

给定两个数字$$a and b$$ ，发现哪个更大$a^b or \, b^a$ 。
如果$a^b > b^a$ ，则打印a ^ b更大
如果$a^b < b^a$ ，打印b ^ a更大
如果$a^b = b^a$ ，打印两者相等
例子：

Input : 3 5
Output : a^b is greater
3^5 = 243, 5^3 = 125. Since, 243>125, therefore a^b > b^a.

Input : 2 4
Output : Both are equal
2^4 = 16, 4^2 = 16. Since, 16=16, therefore a^b = b^a.

蛮力解决方案将是只计算$a^b or \, b^a$并比较它们。但是由于$$a and b$$可以足够大$a^b or \, b^a$即使int long int也无法存储，所以此解决方案不可行。同样计算幂n至少需要$O(logn)$时间使用快速求幂技术。
有效的方法是使用对数。我们必须比较$a^b or \, b^a$ 。如果我们取日志，问题就会减少到比较$$\log_a b \, and \, \log_b a$$ 。
因此，
如果$b\log a > a\log b$ ，则打印a ^ b更大
如果$b\log a < a\log b$ ，打印b ^ a更大
如果$b\log a = a\log b$ ，打印两者相等
下面是上面讨论的有效方法的实现。

C++

// C++ code for finding greater
// between the a^b and b^a
#include 
using namespace std;
 
// Function to find the greater value
void findGreater(int a, int b)
{
    long double x = (long double)a * (long double)(log((long double)(b)));
    long double y = (long double)b * (long double)(log((long double)(a)));
    if (y > x) {
        cout << "a^b is greater" << endl;
    }
    else if (y < x) {
        cout << "b^a is greater" << endl;
    }
    else {
        cout << "Both are equal" << endl;
    }
}
 
// Driver code
int main()
{
    int a = 3, b = 5, c = 2, d = 4;
    findGreater(a, b);
    findGreater(c, d);
    return 0;
}

Java

// Java code for finding greater
// between the a^b and b^a
 
public class GFG{
 
    // Function to find the greater value
    static void findGreater(int a, int b)
    {
        double x = (double)a * (double)(Math.log((double)(b)));
        double y = (double)b * (double)(Math.log((double)(a)));
        if (y > x) {
            System.out.println("a^b is greater") ;
        }
        else if (y < x) {
            System.out.println("b^a is greater") ;
        }
        else {
            System.out.println("Both are equal") ;
        }
    }
     
    // Driver code
    public static void main(String []args)
    {
        int a = 3, b = 5, c = 2, d = 4;
        findGreater(a, b);
        findGreater(c, d);
    }
    // This code is contributed by Ryuga
}

Python 3

# Python 3 code for finding greater
# between the a^b and b^a
import math
 
# Function to find the greater value
def findGreater(a, b):
 
    x = a * (math.log(b));
    y = b * (math.log(a));
    if (y > x):
        print ("a^b is greater");
    elif (y < x):
        print("b^a is greater");
    else :
        print("Both are equal");
 
# Driver code
a = 3;
b = 5;
c = 2;
d = 4;
findGreater(a, b);
findGreater(c, d);
 
# This code is contributed
# by Shivi_Aggarwal

C#

// C# code for finding greater
// between the a^b and b^a
  
using System;
public class GFG{
  
    // Function to find the greater value
    static void findGreater(int a, int b)
    {
        double x = (double)a * (double)(Math.Log((double)(b)));
        double y = (double)b * (double)(Math.Log((double)(a)));
        if (y > x) {
            Console.Write("a^b is greater\n") ;
        }
        else if (y < x) {
            Console.Write("b^a is greater"+"\n") ;
        }
        else {
            Console.Write("Both are equal") ;
        }
    }
      
    // Driver code
    public static void Main()
    {
        int a = 3, b = 5, c = 2, d = 4;
        findGreater(a, b);
        findGreater(c, d);
    }
     
}

PHP

 $x)
    {
        echo "a^b is greater", "\n";
    }
    else if ($y < $x)
    {
        echo "b^a is greater", "\n" ;
    }
    else
    {
        echo "Both are equal", "\n" ;
    }
}
 
// Driver code
$a = 3;
$b = 5;
$c = 2;
$d = 4;
findGreater($a, $b);
findGreater($c, $d);
 
// This code is contributed by ajit
?>

Javascript

// javascript code for finding greater
// between the a^b and b^a
 
    // Function to find the greater value
    function findGreater(a , b) {
        var x =  a * (Math.log( (b)));
        var y =  b * (Math.log( (a)));
        if (y > x) {
            document.write("a^b is greater<br/>");
        } else if (y < x) {
            document.write("b^a is greater<br/>");
        } else {
            document.write("Both are equal<br/>");
        }
    }
 
    // Driver code
     
        var a = 3, b = 5, c = 2, d = 4;
        findGreater(a, b);
        findGreater(c, d);
 
// This code is contributed by todaysgaurav

输出：

a^b is greater
Both are equal

时间复杂度： O(1)

辅助空间： O(1)