09 2019 档案
摘要:# for i in range(100,1000):# if int(str(i)[0])**3+int(str(i)[1])**3+int(str(i)[2])**3==i:# print(i)# else:# continuefor i in range(100,1000): y1=0 for
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摘要:str.marktrans方法:生成一个dict表。对应翻译表(asc码格式),如:t = str.maketrans(‘abcdefghi’, ‘123456789’,“abfgjk”)返回:{97: None, 98: None, 99: 51, 100: 52, 101: 53, 102: N
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