一道填数字题目
- b0207191把0-9这10个数字填入圆圈,使得6条线上的和都一样,
好惭愧,没想出来 - qq149971093想错了
- bt4wang每个节点用了2次
- awk45*2/6=15
- zhdphao简单算了一下,又是玩人的题目。
- b0207191好像没有解,不知道我写得是否有误复制代码
- import numpy as np
- import itertools
- def sigma3(a,b,c):
- return a+b+c
- def sigma4(a,b,c,d):
- return a+b+c+d
- tarray = [0,1,2,3,4,5,6,7,8,9]
- pailie = list(itertools.permutations(tarray))
- countv = 0
- for arr in pailie:
- #print(arr)
- countv+=1
- if countv % 10000 == 0:
- print(countv)
- if sigma4(arr[0],arr[1],arr[2],arr[3])==15 and sigma4(arr[6],arr[7],arr[8],arr[9])==15 and\
- sigma3(arr[0],arr[4],arr[8])==15 and sigma3(arr[1],arr[5],arr[9])==15 and sigma3(arr[2],arr[4],arr[6])==15 and sigma3(arr[3],arr[5],arr[7])==15:
- print(arr)
- break
- import numpy as np
- hollyman楼主其实是来寻大家开心的 iOS fly ~
- 疏导者应该无整数解。
设十个点为:
ABCD
EF
GHIJ
A+B+C+D = G+H+I+J = A+E+I = C+E+G = B+F+J =D+F+H = X
6X = (A+B+C+D) + (G+H+I+J) + (A+E+I) + (C+E+G) + (B+F+J) + (D+F+H)
6X = 2(A+B+C+D+E+F+G+H+I+J) = 2(0+1+2+3+4+5+6+7+8+9) = 90
X = 15
所以每条线的值都是15
令四条斜线相加:
(A+E+I) + (C+E+G) + (B+F+J) + (D+F+H) = 4 X15 = 60
2(E+F) + (A+B+C+D) + (G+H+I+J) = 60
E+F = 15
所以(E,F)可能的取值是{7,8}或者{6,9}。以此出发,剩下的排列组合必然出现矛盾。故没有整数解。