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Suppose we have a number n. We have to check whether we can make express it like a^b or not.

So, if the input is like 125, then the output will be True as 125 = 5^3, so a = 5 and b = 3

To solve this, we will follow these steps −

- if num is same as 1, then:
- return true

- for initialize i := 2, when i * i <= num, update (increase i by 1), do:
- val := log(num) / log(i)
- if val - integer part of val is nearly 0, then:
- return true

- return false

Let us see the following implementation to get better understanding −

#include<iostream> #include<cmath> using namespace std; bool solve(int num) { if (num == 1) return true; for (int i = 2; i * i <= num; i++) { double val = log(num) / log(i); if ((val - (int)val) < 0.00000001) return true; } return false; } int main() { int n = 125; cout << solve(n); }

125

1

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