题目大意: 对于连续的质数$p1$, $p2$, 满足$5 <= p1 <= 1000000$ 求出最小的整数$S$, 它以 $p1$结尾并且能够被$p2$整除. 求$S$的和. 思路: 只需要知道对于一对$p1$, $p2$怎么求对应的$S$.   把$S$表示成$x*10^k+p1$ 其中$k$是$p1$的长度. 然后就转化为求同余方程 $x*10^k+p1\equiv 0\ (mod\ p2)$ 代码: #include <iostream> #include <cst…

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…

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7题:https://projecteuler.net/problem=7 # Project Euler: Problem 7: 10001st prime # By listing the first six prime numbers: # 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. # What is the 10 001st prime number? # Answer…

本题来自 Project Euler 第3题:https://projecteuler.net/problem=3 # Project Euler: Problem 3: Largest prime factor # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number 600851475143 ? # Answer: 6857 a = 6008514751…

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…

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…