The constraints on A and B seem quite large, but the way would be:

Note that (a^b)%c = ((a%c)^b)%c so,

1) Compute (a%c)

2) Use binary exponentiation to compute ((a%c)^b)%c in log(10^100000) time

In python, integers are infinite so u can directly calculate a%c and pow function implements binary exponentiation, so the ans would be simply pow(a%c,b,c).

In C++, where u have no bigint by default, u can input 'a' as a string, and do as follows:

    string a;
    cin>>a;
    ll x,c,ans=0,p=1,n=a.size();
    cin>>c;
    for(int i=n-1;i>=0;--i){
        x=a[i]-'0';
        ans+=x*p%c,ans%=c;
        p=p*10%c;
    

As for binary exponentiation you would need to input b as a string and convert it to a binary string. And then instead of the while(b>0) and b/=2 in the regular binary exponentiation, traverse b in the reverse order replacing the if condition (b%2==1) with b[i]=='1'.

ok.

First the input will be taken as string.
Convery the given numbers in binary then take xor

Iterate through the string and and at each step remainder can be calculated as

public static int power(long x, long y, long p) { long res = 1;

x = x % p;
        if (x == 0)
            return 0;
        while (y > 0)
        {
            if ((y & 1) != 0)
                res = (res * x) % p;

            y = y >> 1;
            x = (x * x) % p;
        }
        return (int)res;
    }

• If $$$B<\phi(C)$$$ then $$$A^B \bmod C = (A \bmod C)^{B} \bmod C$$$

• If $$$B≥\phi(C)$$$ then $$$A^B \bmod C = (A \bmod C)^{\phi(C) + (B \bmod \phi(C))} \bmod C$$$

$$$\phi$$$ here denotes the euler toient function
The above is true for All $$$C$$$

You can calculate :

• $$$\phi(C)$$$ in $$$O(\sqrt C)$$$

• $$$A \bmod C$$$ in $$$O(\log_{10}A)$$$

• $$$B \bmod \phi(C)$$$ in $$$O(\log_{10}B)$$$ with also determining whether $$$B<\phi(C)$$$ or $$$B≥\phi(C)$$$

• $$$A^B \bmod C$$$ in $$$O(\log_{2}\phi(C))$$$ using the new information above