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?編寫函數,功能實返回一個整數的逆序整數。對于輸入的整數n(例如:12345),返回n的逆序整數(54321)。
2019-11-22
源自:初識Python 7-3
正在回答
def square_of_sum(L):
? ? L1=[]
? ? for x in L:
? ? ? ? L1.append(x*x)
? ? return sum(L1)
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
慕妹2214238
def square_of_sum(L):??? x = []??? h = 0??? for l in L:??????? x.append(l * l)??????? h += l*l??? print(x)??? return hprint (square_of_sum([1, 2, 3, 4, 5]))print (square_of_sum([-5, 0, 5, 15, 25]))
#?功能實返回一個整數的逆序整數 #?切片 def?逆轉1(整數): ???????字符化整數?=?str(整數) ???????逆轉后整數?=?int(字符化整數[::-1]) ???????return?逆轉后整數 #字符串拼接????? def?逆轉2(整數): ???????逆轉后整數?=?'' ???????字符化整數?=?str(整數) ???????for?i?in?字符化整數: ???????????逆轉后整數?=?i+逆轉后整數 ???????return?逆轉后整數 ?#還有其他很多辦法,自己研究一下
慕容1173853
def revershow(x=0):
??s=str(x)
????sn=[]
????reverout=''
????for i in range(len(s)):
????????sn.append(s[i:i+1])
????for j in range(len(sn)) :
????????reverout=reverout+sn[len(sn)-j-1]
????print (reverout)
????return
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2019-12-16
def square_of_sum(L):
? ? L1=[]
? ? for x in L:
? ? ? ? L1.append(x*x)
? ? return sum(L1)
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
2019-12-11
def square_of_sum(L):
??? x = []
??? h = 0
??? for l in L:
??????? x.append(l * l)
??????? h += l*l
??? print(x)
??? return h
print (square_of_sum([1, 2, 3, 4, 5]))
print (square_of_sum([-5, 0, 5, 15, 25]))
2019-11-28
2019-11-23
def revershow(x=0):
????sn=[]
????reverout=''
????for i in range(len(s)):
????????sn.append(s[i:i+1])
????for j in range(len(sn)) :
????????reverout=reverout+sn[len(sn)-j-1]
????print (reverout)
????return