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No, C doesn't support any form of overloading (unless you count the fact that the built-in operators are overloaded already, to be a form of overloading).
printf works using a feature called varargs. You make a call that looks like it might be overloaded:
printf("%d", 12); // int overload?
printf("%s", "hi"); // char* overload?
Actually it isn't. There is only one printf function, but the compiler uses a special calling convention to call it, where whatever arguments you provide are put in sequence on the stack[*]. printf (or vprintf) examines the format string and uses that to work out how to read those arguments back. This is why printf isn't type-safe:
char *format = "%d";
printf(format, "hi"); // undefined behaviour, no diagnostic required.
[*] the standard doesn't actually say they're passed on the stack, or mention a stack at all, but that's the natural implementation.
It all depends on how you define "support".
Obviously, C language provides overloaded operators within the core language, since most operators in C have overloaded functionality: you can use binary + with int, long and with pointer types.
Yet at the same time C does not allow you to create your own overloaded functions, and C standard library also has to resort to differently-named functions to be used with different types (like abs, fabs, labs and so on).
In other words, C has some degree of overloading hardcoded into the core language, but neither the standard library nor the users are allowed to do their own overloading.