We define the Perfect Number is a positive integer that is equal to the sum of all its positive divisors except itself.
Now, given an
integer
n, write a function that returns true when it is a perfect number and false when it is not.
Example:
Input: 28
Output: True
Explanation: 28 = 1 + 2 + 4 + 7 + 14
Note: The input number n will not exceed 100,000,000. (1e8)
思路：
- 首先，当然要知道如何得到一个数的divisiors，或者得到一个数是完全数。最简单粗暴的方法，从1到这个数依次看能否被这个数整除。
def checkPerfectNumber(self, num):
"""
:type num: int
:rtype: bool
"""
sum = 0
for i in xrange(1, num):
if (num%i==0):
sum += i
if(sum==num):
return True
else:
return False
但是不幸，在num很大的情况下超时了。
继续分析：28 = 1 + 2 + 4 + 7 + 14
对于2，满足条件，则28/2=14也满足条件，4满足条件，则28/4也满足条件。
所以可以对数字从2到sqrt(num)进行查看，如果满足取模运算为0，则把这个数和num除以这个数的值放到sum里面去。判断是否和num相等即可。
#!/usr/bin/env python
# -*- coding: UTF-8 -*-
class Solution(object):
def checkPerfectNumber(self, num):
"""
:type num: int
:rtype: bool
"""
sum = 0
for i in xrange(1, num):
if (num%i==0):
sum += i
if(sum==num):
return True
else:
return False
def checkPerfectNumber2(self, num):
total, i = 1, 2
while i*i < num:
if num%i==0:
total += i + num/i
i+=1
return num>1 and total==num
def checkPerfectNumber3(self, num):
if num<=1:
return False
sum = cur = 1
while cur<int(num**0.5):
cur+=1
if(num%cur == 0):
sum += cur + num/cur
return sum == num
if __name__ == '__main__':
sol = Solution()
print sol.checkPerfectNumber(6)
print sol.checkPerfectNumber(28)
print sol.checkPerfectNumber(1234)
print sol.checkPerfectNumber2(28)
print sol.checkPerfectNumber3(28)
print sol.checkPerfectNumber3(6)
