data=URLExecute["https://projecteuler.net/problem=13","Table"][[50;;-18]]; num=StringTake[#,50]&/@data[[All,1]]; IntegerDigits[Total[ToExpression/@num]][[;;10]]//FromDigits Problem 14、Longest Collatz sequence
https://projecteuler.net/problem=1projecteuler.net/problem=1后面一段时间用haskell完成,文末有语法介绍。 sumOfMultiples :: (Integral a) => a -> a sumOfMultiples limit = sum [x | x <- [1 .. limit - 1], x `mod` 3 == 0 || x `mod` 5 == 0] sumOfMultiplesImproved :: ...
方法一:最简单的方法为循环1到1000并判断是否能整除3或5 1uintSum=0;2for(uintcurNum=1;curNum<1000;curNum++)if(curNum%3==0|| curNum%5==0)3Sum+=curNum; 执行次数:999次; 方法二:如果每次分别进行相加,再去减掉公倍数 1privateuintCountNumSum(uintbelowNum,uintuIntNum)2{3uintcurSum =0;4fo...
a= 3b= 5total=0foriinrange(1, 1000):ifisMutiple(i, a, b): total+=iprint("Total result is :", total) The solution above needs to iterate everyone from 1 to 1000, Considering that 1 times, 2 times ,3 times of one numbers, we can rewrite it . Here is folllows: defgetCount(...
问尝试解决Project Euler #1的问题EN用python解决 Project Euler 问题 记录 由于比较擅长java 对python...
Project Euler Problem 1: Multiples of 3 and 5,Multiplesof3and5Problem1Ifwelistallthenaturalnumbersbelow10thataremultiplesof3or5,weget3,5,6and9.Thesumo...
Project Euler 笔记 Problem 1 Multiples of 3 or 5 # Find the sum of all the multiples of 3 or 5 below 1000. res = 0 for i in range(1000): if i%3 == 0 or i%5==0: res += i print(res) sum([i for i in range(1000) if (not i%3) or (not i%5)])...
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 ...
Project Euler Problem #11项目 2010/09/21 Greatest product of four adjacent numbers (up, down, left, right, or diagonally) in this 20x20 grid:let grid = [|08;02;22;97;38;15;00;40;00;75;04;05;07;78;52;12;50;77;91;08;
int Ax[9] = { 0,1,2,5,10,20,50,100,200 }, X[9], S = 200; cout << sequncefomula(Ax, X, 8, S); return 0; } ans:73682 第32题: Pandigital products Problem 32 We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once;...