Pandigital multiplesProblem 38Take the number 192 and multiply it by each of 1, 2, and 3:192 × 1 = 192192 × 2 = 384192 × 3 = 576By con...
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2017-03-29 22:01:00
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Pandigital multiplesProblem 38Take the number 192 and multiply it by each of 1, 2, and 3:192 × 1 = 192192 × 2 = 384192 × 3 = 576By concatenating each product we get the 1 to 9 pand
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2022-08-11 16:59:27
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Multiples of 3 and 5If 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.F...
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2015-07-09 14:56:00
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1 --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. 2 --Find the sum of all the multiples of 3 or 5 below 1000. 3 4 --set up iterator 5 function list_iter() 6 local i = 1; 7 local n = 100; 8 return functi...
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2014-03-07 07:22:00
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Pandigital productsProblem 32We shall say that an n-digit number is pandigital if it makes use of all the digits 1 ton exactly once; f...
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2017-03-29 06:38:00
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Pandigital primeProblem 41We shall say that an n-digit number is pandigital if it makes use of all the digits 1 ton exactly once. For ...
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2017-03-28 18:33:00
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Pandigital productsProblem 32We sh
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2022-08-11 15:15:26
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原题链接 考察:构造 思路: 对于n,如果每次x都能+-(n-1)的倍数,那么一定可以使x变成n的倍数. \(x = n*x-(n-1)*x\) \(x+(n-1)x = n*x\) 由此这三步为: 使a[1~n-1]变为n的倍数 使a[n]变成0 使a[1~n]变成0 ##Code #includ
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2021-06-08 02:30:00
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We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.What is the largest n-digit pandigi
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2017-07-28 12:22:00
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Pandigital prime setsProblem 118Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different ...
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2017-03-30 18:52:00
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Pandigital Fibonacci endsProblem 104The Fibonacci sequence is defined by the recurrence relation:Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.It turns out that F541, which contains 113 di
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2022-08-11 14:49:14
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Project Euler 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 the multip
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2013-10-04 11:39:46
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其实不难想,主要看你想的方向对不对思路(想看做法可以直接跳过这部分)\color{Red}(想看做法可以直接跳过这部分)(想看做法可以直接跳过这部分)ai想变成0a_i想变成0ai想变成0一定是最后选择了某个长len得区间,使得自己是len的倍数一定是最后选择了某个长len得区间,使得自己是len的倍数一定是最后选择了某个长len得区间,使得自己是len的倍数但是要在3次操作把n个数变成0,说明这个len区间必须很大很大但是要在3次操作把n个数变成0,说明这个len区间必须很大很大但是要在3次操作
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2021-08-27 09:52:52
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思路：直接离线处理出每个ai 的10倍， 100倍， 1000倍的mod k 后的数值的个数，使用map<int,int >ss[12]存储， ss[x][y]表示 （ai*10x）%k=y的个数，最后处理一下，自己拼接自己的情况。
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2021-07-15 15:04:43
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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=200000+10;
map<int,int> M[11];
int ans[maxn];
short bit[maxn];
int main(){
int n,k,t...
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2022-11-09 18:36:24
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Pandigital Fibonacci endsProblem 104The Fibonacci sequence is defined by the recurrence relation:Fn = Fn−1 + Fn−2, where F1 = 1 and F2...
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2017-03-29 11:38:00
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其实不难想,主要看你想的方向对不对思路(想看做法可以直接跳过这部分)\color{Red}(想看做法可以直接跳过这部分)(想看做法可以直接跳过这部分)ai想变成0a
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2022-02-10 15:36:05
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令$d=\gcd(a,b)$，可以发现$c|(ax+by)$等价于$lcm(c,d)|(ax+by)$，因此不妨令$c'=lcm(c,d)$，然后将$a$、$b$和$c$同时除以$d$ 接下来设$(a,c)=d_{1}$，根据整除的传递性有$d_{1}|(ax+by)$，由于$d_{1}|ax$，可
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2020-12-28 11:00:00
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Multiples of 3 and 5Problem 1
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2022-08-11 15:09:18
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* Solution.php<?php/** * Problem#1 * https://projecteuler.net/problem=1·
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2021-08-15 15:46:24
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