F1,F2=1,2Fk=Fk−1+Fk−2,k=3,4,⋯α,β=1+52,1−52α+β=1,αβ=−1Fk=αk+1−βk+15Fk≈Fk¯=αk+15|Fk¯−Fk|=(β)k+1=(−1α)k+1 假定第k项就是不超过四百万的各项中最后(最大)的项,然后使用定义式中的“递归”技巧 Sk=F2+F4+⋯+FkSk+
问题描述:Project Euler问题#2要求寻找一个数字,使其平方等于2^1024。然而,由于2^1024是一个非常大的数字,目前没有任何已知的数字可以完全平方等于2^1024。因此,这个问题在当前的计算能力下无法得到解决。 名词解释: Project Euler:Project Euler是一个面向数学和编程爱好者的在线数学问题平台,每周会发布一道数学问题,...
int main() { long n = 4000000; long f1, f2 = 1, f3 = 0; long sum = 0; for (f1 = 0; f2 <= n;) { f3 = f2 + f1; if (f3 % 2 == 0) // 如果是偶数 sum += f3; f1 = f2; f2 = f3; } printf("%d", sum ); }...
whileFib[-2]+Fib[-1]<limit: Fib.append(Fib[-2]+Fib[-1]) returnsum(filter(lambdax:x%2==0,Fib)) 方法二: 观察可知斐波那契数列中的奇偶性为 奇偶奇、奇偶奇、奇偶奇,以3为周期循环: 证明A1=1,A2=2,An=An-1+An-2,A3n+2(n为整数)为偶数其他A3n+1,A3n+3(n为整数)为奇数. A1=1,A2...
Project Euler Problem 2: Even Fibonacci numbers Even Fibonacci numbers Problem 2 Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ......
全部来源于Project Euler的题目,持续更新Mathematica解法。Problem 1、 Multiples of 3 and 5Total@Union@Join[Range[3,999,3],Range[5,999,5]] &]Problem 2、 Even Fibonacci numbersbound=NestWhile[#+1&am…
* https://projecteuler.net/problem=2 * Even Fibonacci numbers */ class Solution { protected $bound; public function __construct($bound) { $this->bound = $bound; } public static function fibonacci(int $n) : int { if ($n < 1) {return 0;} ...
Bonus Problems 里的 Problem Secret 触发方式不明,可能是解完一道题以后随机触发 题目 题目本身是一张图 要求是把这张图的像素提取成矩阵,求迭代10^12步后的结果 每步迭代把每个像素替换为上下左右四个像素值的和 最上最下和最左最右视为相邻 最后取模7 ...
Problem 27 Euler discovered the remarkable quadratic formula: n^2+n+41 It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when n=40,402+40+41=40(40+1)+41 is divisible by 41, and certainly when n=41,412+41+41 is clearly divis...
Electron作为一款跨平台的桌面应用端解决方案已经风靡全球。作为开发者，我们几乎不用关心与操作系统的交互...