三个鬼和三个人过河，初始都在一侧，要到另一侧。现在只有一个船，一次做多乘坐两个（人人、人鬼或鬼鬼）。任何时刻任何一侧的鬼的数量多于人，则鬼会把人吃掉，现在要给出一种方案，让鬼和人都过到河的另一侧。未完善。。

fbw.py

# coding=utf-8
'''
Created on 2014-4-27
@author: delfi@lagou.com
'''

def calc_base(base):
    '''calculate the numbers meet the base conditions, l1 and l2 is sorted, then merge sort without repeat'''

    # 0 special
    if base == 0:
        ll = [i * 10 for i in range(1, 11)]
        return ll

    # contain base
    l1 = []
    bsingle = base
    bten = base * 10
    for i in range(1, 10):
        bsingle += 10
        if i != base:
            l1.append(bsingle)
        else:
            l1.extend([bten + j for j in range(10)])
    
    # multiple
    l2 = []
    t = 100 / base + 1
    bmulti = 0
    for i in range(1, t):
        # no use multiplication, use addition for efficient
        bmulti += base
        l2.append(bmulti)
    
    # merge sort, not repeat
    ll = []
    while l1 and l2:
        if l1[0] < l2[0]:
            ll.append(l1.pop(0))
        elif l1[0] > l2[0]:
            ll.append(l2.pop(0))
        else:
            # for not repeat
            l2.pop(0)
    return ll + l1 + l2

def match_condition(n, ll, say, result):
    '''judge whether match the list'''
    for l in ll:
        if n == l:
            result[0] = True
            result[1] = result[1] + say
            break
        elif n < l:
            # because ll is sorted so it can break
            break
    return result

def test_calc():
    '''test the calc_base method'''
    for i in range(10):
        print i, len(calc_base(i))
        print calc_base(i)
        print

if __name__ == '__main__':
    '''I calculate the list by addition rather than use for i in range 100 by division,
     I think the addition is more efficient than division even thought it's like more complex'''
    
    n1 = int(raw_input("Input the first single digits:"))
    n2 = int(raw_input("Input the second single digits:"))
    n3 = int(raw_input("Input the third single digits:"))
    lf = calc_base(n1)
    lb = calc_base(n2)
    lw = calc_base(n3)
    for n in xrange(1, 101):
        result = [False, '']
        match_condition(n, lf, 'Fizz', result)
        match_condition(n, lb, 'Buzz', result)
        match_condition(n, lw, 'Whizz', result)
        if result[0]:
            print result[1]
        else:
            print n

River.py

#coding=utf-8
'''
三个鬼和三个人过河，初始都在一侧，要到另一侧。
现在只有一个船，一次做多乘坐两个（人人、人鬼或鬼鬼）。
任何时刻任何一侧的鬼的数量多于人，则鬼会把人吃掉，现在要给出一种方案，让鬼和人都过到河的另一侧。
Created on 2013-6-28
 
@author: gdh
'''
 
def judge(list):
    '''判断是否被吃'''
    l = len(list)
    gc = 0
    for c in list:
        if c == 'g':
            gc += 1
    if gc > l - gc:
        return False
    return True
 
def success(left, right):
    '''判断是否完成目标'''
    if len(left) == 6 and len(right) == 0:
        return True
    return False
 
def contain(states, left):
    '''判断是否走了重复路'''
    for state in states:
        if len(state) == len(left):
            for c in state:
                if c not in left:
                    return False
            return True    
    return False
 
def back(states, left, right):
    left = states[len(states) - 2]
    left = states[len(states) - 2]
    
def next(states, left, right):
    '''继续下一步'''
    for i in left:
        nleft = left[0, 0]
        nright = right[0, 0]
        
        nright.append(nleft[i])
        del nleft[i]
        if not(judge(nleft) and judge(nright)):
            pass
        
if __name__ == '__main__':
    left = []
    states =[left]
    right = ['r', 'r', 'r', 'g', 'g', 'g']
    while not success(left, right) :
        next(states, left, right)