非黑莫白
发表于 2019-4-17 22:58:38
a=[ for x in range(1,5) for y in range(1,5) for z in range(1,5) if x!=y and y!=z and x!=z]
len(a)
yu123py
发表于 2019-4-26 15:51:48
l = []
for i in range(1, 5):
for j in range(1, 5):
for k in range(1, 5):
if i != j and j!= k and i != k:
l.append(i*100 + j*10 + k)
l = list(set(l))
print(l)
陌小白
发表于 2019-4-29 07:34:40
num_list = []
for i in range(1, 5):
for j in range(1, 5):
if i == j:
continue
for k in range(1, 5):
if j == k or i == k:
continue
num_list.append(i * 100 + j * 10 + k)
print(f'总有{len(num_list)}个这种的数。')
print(f'他们分别是:\n{num_list}')
x287208793
发表于 2019-5-29 15:13:17
number =
for a in number:
for b in number:
for c in number:
if a!=b and a != c and b != c:
print('%d' %(a*100+b*10+c))
糠爸
发表于 2019-7-2 09:02:10
本帖最后由 糠爸 于 2019-7-2 09:15 编辑
count = 0
for a in range(1,5):
for b in range(1,5):
for c in range(1,5):
if a!=b and a!=c and b!=c:
print(a,b,c)
count += 1
print((count)
自己写一遍才有效!
第n个老秃驴
发表于 2019-7-5 16:14:28
n=4*4*4
a=0
while a<4:
a+=1
b=0
while b<4:
b+=1
c=0
while c<4:
c+=1
if a==b or b==c or a==c:
n=n-1
else:
print(100*a+10*b+c)
print ('共有%d种组合'%n)
永恒的蓝色梦想
发表于 2019-8-1 19:32:32
aset={int(str(a)+str(b)+str(c))for a in range(1,5)for b in range(1,5)for c in range(1,5)if a!=b!=c and a!=c}
print(f'结果:{aset}\n个数:{len(aset)}')集合推导
datarec
发表于 2019-8-8 14:31:03
for i in range (1,5):
for j in range (1,5):
for k in range (1,5):
if i != j and i != k and j != k:
print (i,j,k)
这个简单易懂,学习了 16#
inver11
发表于 2019-8-9 15:11:53
本帖最后由 inver11 于 2019-8-9 15:14 编辑
n = 0
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != j and j != k and i != k:
print(i,j,k)
n+=1
print("一共出现了%d次"%n)
inver11
发表于 2019-8-9 15:12:53
清风揽月shine 发表于 2017-4-15 15:10
n=0
for x in range(1,5):
for y in range(1,5):
老哥加了个统计 棒
小香蕉2009
发表于 2019-8-13 15:48:08
本帖最后由 小香蕉2009 于 2019-8-13 16:22 编辑
a = 0
for b in range(1,5):
for c in range(1,5):
for d in range(1,5):
if b != c and c != d and b != d:
a += 1
print(b*100+c*10+d,a)
htygame
发表于 2019-8-23 22:44:23
sums=
for a in sums:
for b in sums:
for c in sums:
if a!=b and b!=c and a!=c:
sumss=a*100+b*10+c
print(sumss)
henry2018
发表于 2019-9-5 17:19:36
i0 = []
for i1 in range(1, 5):
for i2 in range(1, 5):
for i3 in range(1, 5):
i0.append()
print(i0)
当年的小法师
发表于 2019-9-5 22:57:46
n = 0
for h in range(1,5):
for t in range(1,5):
for o in range(1,5):
if h != t and t != o and h != o:
print(100 * h + 10 * t + o)
n += 1
print(n)
lucky邪神
发表于 2019-9-6 02:44:50
counter=0
counter2=0
for x in range(1,5):
for y in range(1,5):
for z in range(1,5):
counter2 +=1
print('组成三位数:%d%d%d'%(x,y,z))
if x!=y and y!=z and x!=z:
counter +=1
print('不重复的数%d%d%d'%(x,y,z))
print('三位数总数是%d,其中%d是不重复的'%(counter2,counter))
轻音少女平泽唯
发表于 2019-9-6 15:01:07
for i in range(1,5):
for j in range((1,5):
for k in range(1,5):
if(i!=k and i!=j and j!=k):
print(i,j,k)
佑爸
发表于 2019-9-24 10:10:51
这要看完第一课才有这个思维?小白表示,不太理解这个,能详细解释下吗,非常感谢
佑爸
发表于 2019-9-24 10:11:25
新手·ing 发表于 2017-3-24 21:43
这是我的解答,欢迎大家一起交流。
这要看完第一课才有这个思维?小白表示,不太理解这个,能详细解释下吗,非常感谢
Freedom911
发表于 2019-10-15 10:53:56
temp =
result = []
for value in temp:
a = int(value / 1 % 10)
b = int(value / 10 % 10)
c = int(value / 100 % 10)
d = int(value / 1000 % 10)
if a + b + c + d == 10 anda * b * c * d == 24:
result.append(value)
print("符合条件是数总共有" + str(len(result)) + "个。他们是:")
for value in result:
print(value)
alina1993
发表于 2019-10-15 22:53:55
本帖最后由 alina1993 于 2019-10-16 11:13 编辑
for z in range (1, 5):
for x in range (1, 5):
for c in range (1, 5):
if z != x != c:
print (z, x, c)
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