cout<<(a^b)%c<<"\n";
What's wrong with the code?
This will not work.
a^b means a xor b in C++.
You have to find a to the power b. Then modulo c.
Refer this
cout<<(a^b)%c<<"\n";
This will not work.
a^b means a xor b in C++.
You have to find a to the power b. Then modulo c.
Refer this