首页 > 其他 > 详细

Problem 6

时间:2019-05-31 21:24:50      阅读:85      评论:0      收藏:0      [点我收藏+]

Problem 6

# Problem_6.py
"""
The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum
找出100一下的所有自然数的平方和与这些自然数的和的平方之间的差数.
"""
sum_of_squares = sum([i*i for i in range(1, 101)])
square_of_sum = pow(sum([i for i in range(1, 101)]), 2)

print(square_of_sum, -, sum_of_squares, =, square_of_sum-sum_of_squares)

 

Problem 6

原文:https://www.cnblogs.com/noonjuan/p/10957316.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!