Arranged probability If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2. The n…

题目链接:点击我打开题目链接 题目大意: 给你一只蚂蚁,它在一个 边长为 \(30-40-50\) 的直角三角形\((x,y)\)上,并且它在直角三角形中选择的位置和移动方向的概率都是相等的.问你这只蚂蚁在最长的边上走出去的概率是多少? 注意:以下是题目的解析. 如果你想要享受做数学题的乐趣就不要看下面了啦.(╯￣Д￣)╯╘═╛ (其实看了也无所谓...) 简要题解: 我们假设蚂蚁在 $ A(x,y) $ 点. 枚举一下题意,画一下图...( 随手画的...丑了点...) 我们可以得到: \(\…

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.…

本题来自 Project Euler 第20题:https://projecteuler.net/problem=20 ''' Project Euler: Problem 20: Factorial digit sum n! means n × (n − 1) × ... × 3 × 2 × 1 For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! i…

本题来自 Project Euler 第17题:https://projecteuler.net/problem=17 ''' Project Euler 17: Number letter counts If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all th…

本题来自 Project Euler 第13题:https://projecteuler.net/problem=13 # Project Euler: Problem 13: Large sum # Work out the first ten digits of the sum of the following one-hundred 50-digit numbers. # Answer: 5537376230 numbers = '''371072875339021027987979982…

本题来自 Project Euler 第4题:https://projecteuler.net/problem=4 # Project Euler: Problem 4: Largest palindrome product # A palindromic number reads the same both ways. # The largest palindrome made from the product # of two 2-digit numbers is 9009 = 91 × 9…

开始做 Project Euler 的练习题.网站上总共有565题,真是个大题库啊! # Project Euler, Problem 1: Multiples of 3 and 5 # If we list all the natural numbers below 10 # that are multiples of 3 or 5, we get 3, 5, 6 and 9. # The sum of these multiples is 23. # Find the sum of all…

题意:三个正整数a + b + c = 1000,a*a + b*b = c*c.求a*b*c. 解法:可以暴力枚举,但是也有数学方法. 首先,a,b,c中肯定有至少一个为偶数,否则和不可能为以上两个等式均不会成立.然后,不可能a,b为奇c为偶,否则a*a%4=1, b*b%4=1, 有(a*a+b*b) %4 = 2,而c*c%4 = 0.也就是说,a和b中至少有一个偶数. 这是勾股数的一个性质,a,b中至少有一个偶数. 然后,解决过程见下(来自project euler的讨论): tag:m…

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…