kankan
想看！！？
import itertools
list1 =
count = 1
for i in itertools.permutations(list1, 4):
list2 = * 9
list2 = 5
if len(set() & set(i)) == 0:
for j in range(4):
list2 = i
list2 = 10 - i
if sum(list2) == 15 and sum(list2) == 15:
print('NO.', count)
for num in range(3):
print(list2)
print()
count += 1
为啥是15啊~想不通 而且504种是包括 {1,2,3}和{1,2,8}这种有重复数字的{:10_285:}
def fun27():
L = []
L2 = []
L3 = []
for i1 in range(1,10):
for i2 in range(1,10):
for i3 in range(1,10):
if i3!=i2 and i2!=i1 and i3!=i1:
L.append(i1)
L.append(i2)
L.append(i3)
for k in range(len(L)//3):
L2.append(L)
for t in range(len(L2)):
L2
for t1 in range(len(L2)):
L2
for t2 in range(len(L2)):
L2
if (((L2+L2+L2)==15) and ((L2+L2+L2)==15)) and (sum(L2)==15) and (sum(L2)==15) and (sum(L2)==15)and ((L2+L2+L2)==15)and ((L2+L2+L2)==15) and ((L2+L2+L2)==15):
L3.append(L2)
L3.append(L2)
L3.append(L2)
print(L3)
L3 = []
结果如下：
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我想问一下，格子里面的数，在三个数组中不能重复出现？
我这个程序的执行效率比较慢
"""
1 2 3
\ | /
4'- + -4
/ | \
3' 2' 1'
由题意得 1, 2, 3, 4 四条路和相等
=> 1 + 1' = 2 + 2' = 3 + 3' = 4 + 4'
从 1 ~ 9 中取四对数，使其和相等，只能首尾配对
(1, 9), (2, 8), (3, 7), (4, 6)
=> 中间为 5
=> 每条路的和为 15
=> 九宫图“四条边”每边和也为 15
从上述四对数中在配对前提下取 3 个数作边，仅能凑 4 组
(1, 6, 8)
(2, 4, 9), (2, 6, 7)
(3, 4, 8)
=> 2, 4, 6, 8 均出现两次，故为四角
2 与 4, 6 搭配，故 2 与 4, 6 相邻，8 为 2 的对角
剩下的数按上述新的四对补齐即可
严格来说，九宫图只有一种解，其他解均由此解或旋转，或镜像得到
题目给出 1 解，旋转可再得 3 解，镜像后又有 4 解 => 共 8 解
"""
def printGridView(nums):
global cnt
cnt += 1
print(f"No.{cnt}:")
for i in range(3):
for j in range(2):
print(nums, end=" | ")
print(nums, "\n- - -")
print('=' * 10)
def mirror(nums):
for i in range(3):
nums, nums = nums, nums
def clockWise(nums):
res = [*3 for _ in range(3)]
for i in range(3):
for j in range(3):
res = nums
return res
def solve(nums):
for i in range(2):
for j in range(4):
nums = clockWise(nums)
printGridView(nums)
mirror(nums)
if __name__ == "__main__":
nums = [,
,
]
cnt = 0
solve(nums)
我感觉太难了，看看答案学习一下。
学习学习
d
1
666
看看
学习
list_num=
times=0
list_sum=[]
#横向相加等于15
for a in list_num:
for b in list_num:
for c in list_num:
if a+b+c==15:
times+=1
list_3=[]
list_3.append(a)
list_3.append(b)
list_3.append(c)
list_sum.append(list_3)
print(a,b,c,times)
print(list_sum)
#纵向第一列相加等于15
list_sum2=[]
for a in list_sum:
for b in list_sum:
for c in list_sum:
list_upright1=[]
if a+b+c==15:
list_upright1.append(a)
list_upright1.append(b)
list_upright1.append(c)
list_sum2.append(list_upright1)
print(list_sum2)
#减去重复含有重复数字的数列
count2=0
count3=0
list_sum3=[]
for a in list_sum2:
count2+=1
print(a,count2)
str1=''
for b in a:
for c in b:
if str(c) not in str1:
str1=str1+str(c)
else:
break
if len(str1)==9:
count3+=1
list_sum3.append(a)
print(list_sum3,count3)
#第二列相加等于15 第三列也自然等于15了
list_sum4=[]
count3=0
for a in list_sum3:
if a+a+a==15:
list_sum4.append(a)
count3+=1
else:
pass
print(list_sum4,count3)
#斜向相加等于15
list_sum5=[]
count4=0
for a in list_sum4:
if a+a+a==15:
list_sum5.append(a)
count4 += 1
else:
pass
print(list_sum5,count4)
list_sum6=[]
count5=0
for a in list_sum5:
if a+a+a==15:
list_sum6.append(a)
count5 += 1
else:
pass
print(list_sum6,count5)
1