hdu 4474 Yet Another Multiple Problem(按位枚举，4级)_are tons of problems about integer multiples. desp-优快云博客

本文介绍了一道关于寻找特定正整数倍数的问题，并提供了一种使用广度优先搜索解决问题的方法。具体而言，给定一个正整数n及一些禁止出现的十进制数字，求最小的n的正倍数，该倍数的十进制表示中不包含这些禁止出现的数字。

Yet Another Multiple Problem

Time Limit: 40000/20000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1754 Accepted Submission(s): 435

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 the following question: Given a positive integer n and m decimal digits, what is the minimal positive multiple of n whose decimal notation does not contain any of the given digits?

Input

There are several test cases.
For each test case, there are two lines. The first line contains two integers n and m (1 ≤ n ≤ 10 ⁴). The second line contains m decimal digits separated by spaces.
Input is terminated by EOF.

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) while Y is the minimal multiple satisfying the above-mentioned conditions or “-1” (without quotation marks) in case there does not exist such a multiple.

Sample Input

Sample Output

  
   Case 1: 2345
Case 2: -1

Source

2012 Asia Chengdu Regional Contest

Recommend

liuyiding

思路：就是数位DP常用的思路，按位枚举。只是这换成了广搜。

模数最多10000 所以最多产生10000个数。

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=5e5+9;
class bit
{
  public:int c,sum,fa;
}Q[mm];
int l,r;
bool ten[11],vis[mm],flag;
int n,m;
void out(int x)
{
  if(Q[x].fa!=-1)
    out(Q[x].fa);
  printf("%d",Q[x].c);
}
void getans()
{  int ret;
     l=0,r=0;flag=0;
    FOR(i,1,9)
    if(!ten[i])
    {
      Q[r].c=i;Q[r].sum=i%n;Q[r].fa=-1;
      ret=Q[r].sum;++r;
      vis[ ret ]=1;
      if(ret==0)
      { flag=1;//puts("yes");cout<<i<<endl;
        out(r-1);return;
      }
    }
    while(l<r)
    {
      FOR(i,0,9)
      if(!ten[i])
      {
        Q[r].c=i;Q[r].sum=(Q[l].sum*10+i)%n;Q[r].fa=l;
        ret=Q[r].sum;
        if(ret==0){out(r);flag=1;return;}
        if(!vis[ ret ]){++r;vis[ret]=1;}
      }
      ++l;
    }
}
int main()
{ int a,cas=0;
  while(~scanf("%d%d",&n,&m))
  { clr(ten,0);
    clr(vis,0);
    printf("Case %d: ",++cas);
    FOR(i,1,m)
    {
      scanf("%d",&a);
      ten[a]=1;
    }
    getans();
    if(flag)puts("");
    else puts("-1");
  }
}