I tried adding following function to check what's going on and the result is

#include"bits/stdc++.h"
using namespace std;
void cmp(double x, int y) {
   cout << setprecision(50) << x << " " << y << " " << x - y << endl;
}
int main()
{
   int n,m;
   cin>>n>>m;
   double a;
   a=log(m)/log(n);
   int b=a;
//   cmp(a, b);
   if(a==b)
   {
       cout<<"YES"<<endl;
       cout<<a-1<<endl;
   }
   else
    cout<<"NO"<<endl;
return 0;
}

When I comment out call of cmp the result is as it was, you get WA, but when adding it all of a sudden you get correct answer (at least on 32 bit machine with g++, on clang you get correct answer) here you have an explanation why sth like this can happen: https://www.linuxtopia.org/online_books/an_introduction_to_gcc/gccintro_70.html, but for uses in cp you don't really need to understand why it happens, what you need to know is that there is some (possibly nondeterministic) function fl such that $$$\frac{|f(x) - x|}{|x|} \leq \nu$$$, where $$$\nu$$$ is roughly $$$10^{-7}$$$ for floats, $$$10^{-16}$$$ for doubles and $$$10^{-19}$$$ for long doubles and whenever you apply any operation involving floating point, e.g. $$$a + b$$$ actual result is $$$fl(a + b)$$$ (except when overflow/underflow occurs)

In particular since any value you will calculate will be passed through $$$fl$$$ function multiple times comparing floats with == will hardly ever return true (and by != hardly ever return false). If you want to compare two double values do $$$abs(a - b) \leq \varepsilon$$$ for some small value of $$$\varepsilon$$$ (checking if something is an integer is the same as checking if it is equal to llround(x)). When the calculation is so short as in this problem eps = 1e-10 is usually fine, but if we want to be more rigorous we need to estimate how close to an integer a number of form $$$\log_{a}(b)$$$ with $$$2\leq a, b < 2^{31}$$$ be if it's not an integer.

$$$\frac{\log{a}}{\log{b}} - k = \frac{\log{a} - k\log{b}}{\log b} = \frac{\log{a} - \log{b^k}}{\log b}$$$ By Lagrange mean value theorem there is $$$\xi$$$ in $$$[a, b^k]$$$ (or $$$[b^k, a]$$$) such that $$$\log{a} - \log{b^k} = \log'(\xi)(a - b^k) = \frac{(a - b^k)}{\xi}$$$. Numerator is an integer and we assumed it's not $$$0$$$, so it's at least $$$1$$$ by absolute value, denominator is at most $$$2^31 \approx 2\cdot 10^9$$$, so we're off by at least $$$5\cdot10^{-10}$$$ divided by $$$\log(\xi)$$$, which is at least $$$\log(2) \approx 0.69$$$ (but of course assuming it's $$$\approx 1$$$ is good enough), so $$$|\frac{\log{a}}{\log{b}} - k| \geq \frac{1}{2\cdot10^9\log{2}} > 10^{-10}$$$, value we will actually calculate is off by roughly $$$10^{-19}$$$ if we use long doubles, so picking any epsilon in between will be fine. If on the other hand it turned out actual value can be off by e.g. $$$10^{-24}$$$ it would mean floating points aren't enough to determine if result is an integer and we need to find some different solution