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 nat…

我的做法就是暴力,1+...+n 用前n项和公式就行 1^2+2^2+....+n^2就暴力了 做完后在讨论版发现两个有趣的东西. 一个是 (1+2+3+...+n)^2=(1^3)+(2^3)+(3^3)+...+(n^3) 另一个是 1^2+2^2+....+n^2的公式以及推导. 假设f(n) = an3 + bn2 + cn + d 已知:f(0) = 0; f(1) = 1; f(2) = 5; f(3) = 14 然后带入解方程,得到a,b,c,d,再用归纳法证明其正确性. 证明过程…

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.…

简单: e 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 n…

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 nat…

In Problem 42 we dealt with triangular problems, in Problem 44 of Project Euler we deal with pentagonal number, I can only wonder if we have to deal with septagonal numbers in Problem 46. Anyway the problem reads Pentagonal numbers are generated by t…

It is possible to show that the square root of two can be expressed as an infinite continued fraction.  2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... By expanding this for the first four iterations, we get: 1 + 1/2 = 3/2 = 1.51 + 1/(2 + 1/2) = 7…

基础的动态规划...数塔取数问题. 状态转移方程: dp[i][j] = num[i][j] + max(dp[i+1][j],dp[i+1][j+1]);…

直接python搞过.没啥好办法.看了下别人做的,多数也是大数乘法搞过. 如果用大数做的话,c++写的话,fft优化大数乘法,然后快速幂一下就好了.…

Summation of primes Problem 10 The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. The code resemble : import math limit = 2000000 crosslimit = int(math.sqrt(limit)) #sieve = [False] * limit sieve =…