代码拉取完成,页面将自动刷新
#Way2 1:
#isHappyNumber() will determine whether a number is happy or not
def isHappyNumber(num):
rem = sum = 0;
#Calculates the sum of squares of digits
while(num > 0):
rem = num%10;
sum = sum + (rem*rem);
num = num//10;
return sum;
num = 82;
result = num;
while(result != 1 and result != 4):
result = isHappyNumber(result);
#Happy number always ends with 1
if(result == 1):
print(str(num) + " is a happy number after apply way 1");
#Unhappy number ends in a cycle of repeating numbers which contain 4
elif(result == 4):
print(str(num) + " is not a happy number after apply way 1");
#way 2:
#Another way to do this and code is also less
n=num
setData=set() #set datastructure for checking a number is repeated or not.
while 1:
if n==1:
print("{} is a happy number after apply way 2".format(num))
break
if n in setData:
print("{} is Not a happy number after apply way 2".format(num))
break
else:
setData.add(n) #adding into set if not inside set
n=int(''.join(str(sum([int(i)**2 for i in str(n)])))) #Pythonic way
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。