Almost equilateral triangles It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the almost equilateral triangle 5-5-6 has an area of 12 square units. We shall define an almost equilateral tr…

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…

Right triangles with integer coordinates The points P (x1, y1) and Q (x2, y2) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ. There are exactly fourteen triangles containing a right angle that can be formed whe…

传送门 做出一个好几个星期屯下来的题目的感觉就是一个字: 爽! 上图的黄点部分就是我们需要求的点 两边的部分很好算 求圆的地方有一个优化,由于圆心是整数点,我们可以把圆分为下面几个部分,阴影部分最难算,最后乘就好了 代码如下所示 #include <bits/stdc++.h> using namespace std; const int MAXN = 2005; const int INF = 0x3f3f3f3f; typedef long long ll; typedef long do…

题意:若三边长 { a , b , c } 均为整数的直角三角形周长为 p ,当 p = 120 时,恰好存在三个不同的解:{ 20 , 48 , 52 } , { 24 , 45 , 51 } , { 30 , 40 , 50 } 在所有的p ≤ 1000中,p取何值时有解的数目最多? 思路:可以构建素勾股数,每构建成功一组素勾股数就用其生成其他勾股数,最后扫描一遍取最大值即可. 素勾股数性质: /************************************************…

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 ×…

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.…

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…