a(T):rsrrsr b(T):a(2-T)b(T-5) c(T):b(T)sc(T-1) c(20)
In this code, b(T) equals "rsrrsr" if T≡1 mod 5, and no moves otherwise.
f(T):sf(T-4)rf(T+7) f(8)
This code realizes the loop
a:ssrsssrsssrsssr a
In this loop, the number of "s" equals 4 + 7, and the number of "r" equals 7.
In generally, the code
f(T):[1]f(T-A)[2]f(T+B)
implies a loop with (A+B) times of "[1]" and B times of "[2]".
a(T):sa(T-37)ra(T+83) a(1)
Since 83/37 approximately equals 9/4, this code is similar to
a(T):sa(T-4)ra(T+9) a(1)
but the small difference changes the direction of the movement.
Therefore, "square minus square" shape can be drawn by a 12B code.
係数が1じゃないやつ
f(T):sf(T-4)rf(T-7) f(255)
Probably, no problems in HOJ requires this method (even for "best solution") at this moment.
a(X):XXXXrXXXXla(srslX) a(r) (20B)
The code above can be compressed by replacing XXXX by another variable.
a(X,Y):YrYla(srslX,XXXX) a(r,) (19B)
Zとか?
f(X):ssXlssf(sX) f() (12B)
This code can be compressed by recycling "ss":
f(X):ssXlf(f(sX)) f() (11B)
育たない再帰とかで