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…

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…

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…

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]); } }…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4474 (a*10+b)%c = ((a%c)*10+b%c)%c; 然后从高位开始枚举能填的数字填充,只是注意最高位(第一位)不能为0. 代码: #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<string> #include<…

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…

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 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…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53016    Accepted Submission(s): 20171 Problem Description The least common multiple (LCM) of a set of positive integers is…