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…

[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…

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…

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[][…

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…

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…

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…

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…

[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.…

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…