因此就出現了一個難題:的Prolog:如何優化這個代碼(求解123456789 = 100拼圖)
這個方程是不完整的:1 2 3 4 5 6 7 8 9 = 100.一個方法,使 它準確的是通過添加七個加號和減號,如下所示:1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 = 100. 如何使用只有3加號或減號?
我是很新的序言,解決了這個難題,但我不知道如何去優化它
makeInt(S,F,FinInt):-
getInt(S,F,0,FinInt).
getInt(Start, Finish, Acc, FinInt):-
0 =< Finish - Start,
NewAcc is Acc*10 + Start,
NewStart is Start +1,
getInt(NewStart, Finish, NewAcc, FinInt).
getInt(Start, Finish, A, A):-
0 > Finish - Start.
itCounts(X,Y,Z,Q):-
member(XLastDigit,[1,2,3,4,5,6]),
FromY is XLastDigit+1,
numlist(FromY, 7, ListYLastDigit),
member(YLastDigit, ListYLastDigit),
FromZ is YLastDigit+1,
numlist(FromZ, 8, ListZLastDigit),
member(ZLastDigit,ListZLastDigit),
FromQ is ZLastDigit+1,
member(YSign,[-1,1]),
member(ZSign,[-1,1]),
member(QSign,[-1,1]),
0 is XLastDigit + YSign*YLastDigit + ZSign*ZLastDigit + QSign*9,
makeInt(1, XLastDigit, FirstNumber),
makeInt(FromY, YLastDigit, SecondNumber),
makeInt(FromZ, ZLastDigit, ThirdNumber),
makeInt(FromQ, 9, FourthNumber),
X is FirstNumber,
Y is YSign*SecondNumber,
Z is ZSign*ThirdNumber,
Q is QSign*FourthNumber,
100 =:= X + Y + Z + Q.