Highly divisible triangular number Problem 12 The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36,…

最直接的想法就是暴力搞搞,直接枚举,暴力分解因子.再好一点,就打个素数表来分解因子.假设num=p1^a1p2^a2...pn^an,则所有因子个数为(a1+1)(a2+1)...(an+1). 再好一点呢,还是要利用积性函数.假设因子个数函数为f(x),因为num=n(n+1)/2,所以num2=n(n+1),因为n和n+1为相邻的正整数,所以互质,所以f(2num)=f(n)f(n+1), 则f(num)=f(n)f((n+1)/2)(n+1为偶数)或者f(num)=f(n/2)*f(n+1…

我的那个暴力求解,太耗时间了. 用了网上产的什么因式分解,质因数之类的.确实快!还是数学基础不行,只能知道大约. The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 2…

title: The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... Let us list the factors of…

本题来自 Project Euler 第12题:https://projecteuler.net/problem=12 # Project Euler: Problem 12: Highly divisible triangular number # The sequence of triangle numbers is generated by adding the natural numbers. # So the 7th triangle number would be 1 + 2 + 3…

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…

本题来自 Project Euler 第21题:https://projecteuler.net/problem=21 ''' Project Euler: Problem 21: Amicable numbers Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a ≠ b…

本题来自 Project Euler 第11题:https://projecteuler.net/problem=11 # Project Euler: Problem 10: Largest product in a grid # In the 20×20 grid below, four numbers along a diagonal line have been marked in red. # The product of these numbers is 26 × 63 × 78 ×…

本题来自 Project Euler 第5题:https://projecteuler.net/problem=5 # Project Euler: Problem 5: Smallest multiple # 2520 is the smallest number that can be divided by # each of the numbers from 1 to 10 without any remainder. # What is the smallest positive num…

上一次接触 project euler 还是2011年的事情,做了前三道题,后来被第四题卡住了,前面几题的代码也没有保留下来. 今天试着暴力破解了一下,代码如下: (我大概是第 172,719 个解出这道题的人) program 4 A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.…