1. 题目：
In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants will be placed in a large circle, facing inwards. Someone is arbitrarily chosen as number 1, and the rest are numbered counter-clockwise up to N (who will be standing on 1’s left). Starting from 1 and moving counter-clockwise, one labour official counts off k applicants, while another official starts from N and moves clockwise, counting m applicants. The two who are chosen are then sent off for retraining; if both officials pick the same person she (he) is sent off to become a politician. Each official then starts counting again at the next available person and the process continues until no-one is left. Note that the two victims (sorry, trainees) leave the ring simultaneously, so it is possible for one official to count a person already selected by the other official.
Input
Write a program that will successively read in (in that order) the three numbers (N, k and m; k; m>0, 0<N<20) and determine the order in which the applicants are sent off for retraining. Each set ofthree numbers will be on a separate line and the end of data will be signalled by three zeroes (0 0 0).Output
For each triplet, output a single line of numbers specifying the order in which people are chosen. Each number should be in a field of 3 characters. For pairs of numbers list the person chosen by the counter-clockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).
Note:The symbol ⊔ in the Sample Output below represents a space.Sample Input
10 4 3
0 0 0Sample Output
␣␣4␣␣8,␣␣9␣␣5,␣␣3␣␣1,␣␣2␣␣6,␣10,␣␣7
2. 要点：
- 函数在主函数中调用必须先定义后才能调用，当被其它函数调用时可以后定义
- 起始点的选择：n-1, 0
- 将选中的元素设为0，其后遍历序列时逢0跳过，可减免对遍历步数的复杂判断
- 取余实现循环。负数取余：（-5）%3 = 1
3. 代码：
#! python3
def go(queue,p,d,t):
while t:
p = (p + d)%len(queue)
if queue[p] != 0:
t -= 1
return p
n, k, m = [int(i) for i in input().split()]
results = []
while n:
queue = [i+1 for i in range(n)]
result = []
left = n
p1, p2 = n-1, 0
while left:
p1 = go(queue,p1,1,k)
p2 = go(queue,p2,-1,m)
if p1 == p2:
result.append('%3s' % (str(queue[p1])))
left -= 1
else:
result.append('%3s%3s' % (queue[p1],queue[p2]))
left -= 2
queue[p1] = queue[p2] = 0
results.append(','.join(result))
n, k, m = [int(i) for i in input().split()]
for i in results:
print(i)
