Project Euler 66: Diophantine equation】的更多相关文章

题目链接 思路: 连分数求佩尔方程最小特解 参考博客 模板: LL a[]; bool min_pell(LL d, LL &x, LL &y) { LL m = floor(sqrt(d+0.5)); if(m*m == d) return false; ; a[cnt++] = m; LL b = m, c = ; double sq = sqrt(d); do { c = (d - b*b)/c; a[cnt++] = (LL)floor((sq+b)/c); b = a[cnt-]…
Diophantine reciprocals II In the following equation x, y, and n are positive integers. For n = 4 there are exactly three distinct solutions: It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n for…
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…
project euler 169 题目链接:https://projecteuler.net/problem=169 参考题解:http://tieba.baidu.com/p/2738022069 #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define mp make_pair #define pb push_back #define rep(i, a, b) for(i…
题目要求是: The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832. 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319…
Project Euler: 欧拉计划是一系列挑战数学或者计算机编程问题,解决这些问题需要的不仅仅是数学功底. 启动这一项目的目的在于,为乐于探索的人提供一个钻研其他领域并且学习新知识的平台,将这一平台打造一个有趣和休闲 的环境. 项目主页:https://projecteuler.net 第一题 Multiples of 3 and 5 If we list all the natural numbers below 10 that are multiples of 3 or 5, we ge…