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发表于 2016-9-20 17:36:25
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- 2 6 6 2 5 1 2 1 1 0
- 2783915460
- [Finished in 3.0s]
复制代码
- count = 0
- for a in range(10):
- for b in range(9):
- for c in range(8):
- for d in range(7):
- for e in range(6):
- for f in range(5):
- for g in range(4):
- for h in range(3):
- for i in range(2):
- for j in range(1):
- count += 1
- if count == 1000000:
- print (a,b,c,d,e,f,g,h,i,j)
- a1,b1,c1,d1,e1,f1,g1,h1,i1,j1 = a,b,c,d,e,f,g,h,i,j
- lst = [0,1,2,3,4,5,6,7,8,9]
- final = str(lst.pop(a1))
- final += str(lst.pop(b1))
- final += str(lst.pop(c1))
- final += str(lst.pop(d1))
- final += str(lst.pop(e1))
- final += str(lst.pop(f1))
- final += str(lst.pop(g1))
- final += str(lst.pop(h1))
- final += str(lst.pop(i1))
- final += str(lst.pop(j1))
- print (final)
复制代码
虽然代码丑,但是好处是速度快,3秒出结果。
第一行表示的是每个数字变动的次数。
第二行就根据每个数字变动的次数计算答案。 |
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