D. Beautiful numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if…

[codeforces 55]D. Beautiful numbers 试题描述 Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and…

D. Beautiful numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if…

C. Beautiful Numbers 题目连接: http://www.codeforces.com/contest/300/problem/C Description Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal representation of this integer only contains…

D. Beautiful numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/55/problem/D Description Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only i…

题意: 如果一个数能被自己各个位的数字整除,那么它就叫 Beautiful numbers.求区间 [a,b] 中 Beautiful numbers 的个数. 分析:先分析出,2~9 的最大的最小公倍数是 2520({5,7,8,9}),先预处理出所有可能的最小公倍数m[c] dp[i][d][c]表示长度i, 余数d,各位上的数的最小公倍数是m[c]的个数. #include<cstdio> #include<cstring> #define mod 2520 ll dp[][…

D. Beautiful numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if…

链接:https://ac.nowcoder.com/acm/problem/17385来源:牛客网 题目描述 NIBGNAUK is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by the sum of its digits.   We will not argue…

传送门 参考资料: [1]:CodeForces 55D Beautiful numbers(数位dp&&离散化) 我的理解: 起初,我先定义一个三维数组 dp[ i ][ j ][ k ]:来到 i 位置时,所有非零数的lcm = j,当前数位 k 时含有的 Beautiful numbers 的个数. 但是,由题意得,当前的数 k 可以是个很大的数(9e18),数组根本就开不下,那怎么办呢? 将当前的数 hash 一下,如何hash呢? 假设 a,b,c,d 为[0,9]的数,那么不存…

目录 Beautiful Numbers PROBLEM 题目描述 输入描述: 输出描述: 输入 输出 MEANING SOLUTION CODE Beautiful Numbers PROBLEM 题目描述 NIBGNAUK is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisib…

CF55D Beautiful numbers 题意 \(t(\le 10)\)次询问区间\([l,r](1\le l\le r\le 9\times 10^{18})\)中能被每一位上数整除的数的个数,0算可以整除任何数. 最开始写了个假做法,既没算0复杂度也是假的 最开始把每个模数都压进去了,后来发现只压2520就可以了 然后前两位压前\(i\)位与\(\bmod 2520\)的结果,第三位最开始压了每个数字出现集合,这样就很屑. 可以直接压数字集合的最小公倍数,仅有不到50个取值,做一个H…

D. Beautiful numbers 链接 题意: 求[L,R]中多少个数字可以整除它们的每一位上的数字. 分析: 要求模一些数字等于0等价于模它们的lcm等于0,所以可以记录当前出现的数字的lcm,最后判断组成的数字是否模lcm等于0. 但是这个数字太大记录不下.根据一个性质a%b=(a%kb)%b,所以可以记录当前的数字模2520的值,最后模一下lcm. 这样的状态是$20 \times 2520 \times 2520$的,状态太大了,考虑如何缩小空间.因为1~9的lcm只能是50个左…

[CF55D]Beautiful numbers(动态规划) 题面 洛谷 CF 题解 数位\(dp\) 如果当前数能够被它所有数位整除,意味着它能够被所有数位的\(lcm\)整除. 所以\(dp\)的时候前面所有数的\(lcm\)要压进\(dp\)值中. 又因为\(lcm\)的余数也是有意义的,但是又不能暴力记, 所以记录一下\([1,9]\)所有数的\(lcm\)也就是\(2520\)就好了. 但是数组太大,实际上,有意义的\(lcm\)个数只有不到\(50\)个,重新编号就可以压缩状态了.…

[Codeforces-div.1 55D] Beautiful numbers 试题分析 还是离散化...\(f_{i,j,k}\)表示i位,gcd为j,余数为k. #include<iostream> #include<cstring> #include<cstdio> #include<vector> #include<algorithm> using namespace std; #define LL long long inline L…

C. Beautiful Numbers Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 524288/262144K (Java/Other) Total Submission(s) : 27   Accepted Submission(s) : 7 Problem Description Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly…

题目链接:http://codeforces.com/problemset/problem/55/D D. Beautiful numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Volodya is an odd boy and his taste is strange as well. It seems to…

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful num…

题目链接:https://vjudge.net/problem/CodeForces-55D D. Beautiful numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Volodya is an odd boy and his taste is strange as well. It seems to him…

C. Beautiful Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal…

http://codeforces.com/problemset/problem/55/D Beautiful Numbers : 这个数能整除它的全部位上非零整数.问[l,r]之间的Beautiful Numbers的个数. 若一个数能整除它的全部的非零数位.那么相当于它能整除个位数的最小公倍数. 因此记忆化搜索中的參数除了len(当前位)和up(是否达到上界),有一个prelcm表示前面的数的最小公倍数.推断这个数是否是Beautiful Numbers,还要有一个參数表示前面数,可是这个数…

A - Beautiful numbers Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the qua…

[CF55D]Beautiful numbers 题面 洛谷 题解 考虑到如果一个数整除所有数那么可以整除他们的\(lcm\),而如果数\(x\)满足\(x\bmod Lcm(1,2...,9)=r\),且\(r\bmod Lcm\{x有的数\}=0\),那么这个数一定满足条件. 因为\(Lcm(1,2...,9)=2520\)比较小,所以我们可以存下来. 考虑数位dp,我们设\(f[i][lcm][r]\)表示目前\(dp\)到第\(i\)位,当前已选的数的\(lcm\)为\(lcm\),前面…

链接: https://vjudge.net/problem/CodeForces-55D 题意: Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with…

2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 J Beautiful Numbers (数位DP) 链接:https://ac.nowcoder.com/acm/contest/163/J?&headNav=acm来源:牛客网 时间限制:C/C++ 8秒,其他语言16秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 NIBGNAUK is an odd boy and his taste is strange a…

Codeforces 55D. Beautiful numbers 题意 求[L,R]区间内有多少个数满足:该数能被其每一位数字都整除(如12,24,15等). 思路 一开始以为是数位DP的水题,觉得只需要记录搜到当前位出现了哪些数字作为状态即可,明显是假算法...感觉这是一道数位DP好题.可以这样思考:一个数要想被其各位数字分别都整除,等价于它被那些数字的LCM整除.因此记录当前位,当前数对(1~9的LCM)取模的结果,当前出现的数字的LCM这三个值作为状态才合理,即dp[pos][sum][…

来源:http://codeforces.com/problemset/problem/1265/B   B. Beautiful Numbers   You are given a permutation p=[p1,p2,…,pn]p=[p1,p2,…,pn] of integers from 11 to nn . Let's call the number mm (1≤m≤n1≤m≤n ) beautiful, if there exists two indices l,rl,r (1≤l…

题目 Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful…

题目 Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful…

题目找链接 题意: 如果数a能被a中的每一位数整除(0除掉),则称a是一个beautiful number,求一个区间内的beautiful numbers的个数. 分析: 首先,很显然,l到r的所有beautiful numbers(以下写b n吧,有点长)等于0到r的b n个数减去0到l-1的b n的个数,这个不用证明吧.当然也很好证.然后我们只要能在时间允许的范围内求0到某个数的b n的数量就可以了. 怎么求能?我们把它转化到树上,先举个例子:如果要求0到23的b n的数量,那么我们假设有…

链接: https://codeforces.com/contest/1265/problem/B 题意: You are given a permutation p=[p1,p2,-,pn] of integers from 1 to n. Let's call the number m (1≤m≤n) beautiful, if there exists two indices l,r (1≤l≤r≤n), such that the numbers [pl,pl+1,-,pr] is a…