递归:
def J(n,x): return 0 if n==1 else (J(n-1,x)+x-1) % n
循环:
def J(n,x): k=0 for i in range(2,n+1): k=(k+x)%i return k+1
列表推倒:
def J(n,x): return reduce(lambda t, _: (t[(x-1) % len(t):] + t[:(x-1) % len(t)])[1:], range(n-1), range(1, n+1))
def J(n,x): li=range(1,n+1) while li: print '\t'.join('1' if i in li else '0' for i in range(1,n+1)) li=(lambda t: li[t:]+li[:t])( (x-1)%len(li) )[1:]J(10, 5) #Out:#1 1 1 1 1 1 1 1 1 1#1 1 1 1 0 1 1 1 1 1#1 1 1 1 0 1 1 1 1 0#1 1 1 1 0 0 1 1 1 0#1 0 1 1 0 0 1 1 1 0#1 0 1 1 0 0 1 1 0 0#1 0 1 1 0 0 1 0 0 0#0 0 1 1 0 0 1 0 0 0#0 0 1 0 0 0 1 0 0 0#0 0 1 0 0 0 0 0 0 0
In Mathematica :
