K - Yet Another Multiple Problem Time Limit:20000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4474 Appoint description:  System Crawler  (2014-10-16) Description There are tons of problems about integer mul…

Yet Another Multiple Problem Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3407    Accepted Submission(s): 825 Problem Description There are tons of problems about integer multiples. Despit…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. Fo…

Yet Another Multiple Problem Time Limit : 40000/20000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other) Total Submission(s) : 2   Accepted Submission(s) : 1 Problem Description There are tons of problems about integer multiples. Despite the f…

Yet Another Multiple Problem Description There are tons of problems about integer multiples. Despite the fact that the topic is not original, the content is highly challenging. That’s why we call it “Yet Another Multiple Problem”. In this problem, yo…

Problem Description There are tons of problems about integer multiples. Despite the fact that the topic is not original, the content is highly challenging. That’s why we call it “Yet Another Multiple Problem”.In this problem, you’re asked to solve th…

题意很简单,要你用一些数字,组成一个数的倍数,且那个数最小. 比赛的时候没能做出来,深坑啊. 其实我只想说我以前就做过这种类型的题目了,诶. 题目的解法是数位宽搜. 首先把可用的数位提取出来,从小到大一个一个放到后面去.这样保证了出现的数字一定是最小的. 同时记录出现过的余数的情况,因为同一个余数k*10+ai再对n取余的数是相同的,所以对于同一个余数我们不需要走两遍,这样就保证了搜索的时间复杂度为O(n). 我们只要对于每一个状态,直接枚举可选择的数字放在他们后面,直到找到一个余数为0的就可以…

题意:求m的倍数中不包含一些数码的最小倍数数码是多少.比如15 ,不包含0  1 3,答案是45. BFS过程:用b[]记录可用的数码.设一棵树,树根为-1.树根的孩子是所有可用的数码,孩子的孩子也是所有可用的数码.这样从根到叶子节点这条路径所组成的数表示一个可行的数. __                 __ 剪枝:(A % m ==  B % m)  =>  (AX % m ==  BX % m)   即如果搜索到一个数X, X%m == a (a !=0) , 则以后如果搜索到Y , Y…

题意: 找到一个n的倍数,这个数不能含有m个后续数字中的任何一个 题解: #include<stdio.h> #include<string.h> #include<queue> using namespace std; queue<int>que; ]; ],pre[],value[]; int n,m; void print(int n) { ) { print(pre[n]); printf("%d",value[n]); } }…

没什么巧办法,直接搜就行. 用余数作为每个节点的哈希值. #include <cstdio> #include <cstring> #include <cstdlib> ; struct node { int mod; int fa; int digit; node() {} node( int mod, int fa, int dig ):mod(mod), fa(fa), digit(dig) { } }; int N; ]; bool vis[MAXN]; nod…