(RESENJA) Zadaci programiranje 11/04/2016

st10(N, 1):- N < 10.
st10(N, R):- N > 9, N1 is N div 10, st10(N1, R1), R is R1 * 10.

/* Prvi */
p2(N):- N mod 10 =:= 2.
p2(N):- N > 0, N1 is N div 10, p2(N1).

/* Drugi */
sqr(1, A, R):- R is A.
sqr(N, A, R):- N > 1, N1 is N - 1, sqr(N1, A, R1), R is R1 * A.

/* Treci */
z3(N, N):- N < 10, N mod 2 =:= 1.
z3(N, R):- N < 10, R is N div 2.
z3(N, R):- N > 9, (N mod 10) mod 2 =:= 0, N1 is N div 10, z3(N1, R1), R is R1 * 10 + (N mod 10) div 2.
z3(N, R):- N > 9, (N mod 10) mod 2 =:= 1, N1 is N div 10, z3(N1, R1), R is R1 * 10 + (N mod 10).

/* Cetvrti */
z4(N, N):- N < 10, N =\= 1.
z4(N, R):- N > 9, N mod 10 =:= 1, N1 is N div 10, z4(N1, R).
z4(N, R):- N > 9, N mod 10 =\= 1, N1 is N div 10, z4(N1, R1), R is (R1 * 10) + (N mod 10).
z4(N, 0):- N < 10, N =:= 1.

/* Peti */
inv(N, R):- N < 10, R is N.
inv(N, R):- N > 9, st10(N, ST10), N1 is N mod ST10, inv(N1, R1), R is (R1 * 10) + (N div ST10).

/* Peti optimalan */
inverz(N, I):- inverz(N, I, 0).
inverz(N, I, A):- N > 0, N1 is N mod 10, A1 is A * 10 + N1,
    N2 is N div 10, inverz(N2, I, A1).
inverz(0, A, A).