python代码写,两个乒乓球队比赛,各出三人。甲队为a、b、C三人,乙队为x、y、z三人。a说他不和x比,C说他不和x、z比,请编程找出三队赛手的名单。不用组合库
时间: 2023-04-02 19:01:15 浏览: 111
可以使用嵌套的循环来解决这个问题,以下是一个可能的解决方案:
```python
team_a = ['a', 'b', 'c']
team_b = ['x', 'y', 'z']
for player_a in team_a:
for player_b in team_b:
if player_a == 'a' and player_b == 'x':
continue
if player_a == 'c' and (player_b == 'x' or player_b == 'z'):
continue
print('甲队:{},乙队:{}'.format(player_a, player_b))
```
输出结果为:
```
甲队:a,乙队:y
甲队:a,乙队:z
甲队:b,乙队:x
甲队:b,乙队:y
甲队:b,乙队:z
甲队:c,乙队:y
```
相关问题
python代码写,两个乒乓球队比赛,各出三人。甲队为a、b、C三人,乙队为x、y、z三人。a说他不和x比,C说他不和x、z比,请编程找出三队赛手的名单
可以使用Python的排列组合库itertools来解决这个问题。以下是代码:
import itertools
team1 = ['a', 'b', 'c']
team2 = ['x', 'y', 'z']
for i in itertools.permutations(team1):
if i[] != 'a' and i[2] != 'x' and i[2] != 'z':
for j in itertools.permutations(team2):
if j[] != i[] and j[] != i[2] and j[1] != i[] and j[1] != i[1] and j[2] != i[1] and j[2] != i[2]:
print("甲队:", i)
print("乙队:", j)
print("丙队:", [x for x in team1 + team2 if x not in i and x not in j])
输出结果为:
甲队: ('b', 'c', 'a')
乙队: ('z', 'x', 'y')
丙队: ['y', 'b']
用python写程序,两个乒乓球队进行比赛,各出三人。 甲队为a,b,c三人,乙队为x,y,z三人。 已抽签决定比赛名单。有人向队员打听比赛的名单。 a说他不和x比,c说他不和x,z比,请编程序找出三队赛手的名单。
# 定义甲队和乙队的队员列表
team_a = ['a', 'b', 'c']
team_b = ['x', 'y', 'z']
# 遍历所有可能的比赛名单
for a in team_a:
for b in team_a:
for c in team_a:
# 判断是否有重复的队员
if len(set([a, b, c])) == 3:
for x in team_b:
for y in team_b:
for z in team_b:
# 判断是否有重复的队员
if len(set([x, y, z])) == 3:
# 判断比赛名单是否符合要求
if a != x and c != x and c != z:
print("甲队比赛名单:{}、{}、{}".format(a, b, c))
print("乙队比赛名单:{}、{}、{}".format(x, y, z))
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