hdu 4474 Yet Another Multiple Problem BFS+同余判重 ★★★★

Yet Another Multiple Problem

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4947    Accepted Submission(s): 1082

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 ≤ 10⁴). 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

2345 37 8 9100 10

 

Sample Output

Case 1: 2345Case 2: -1

 

Source

2012 Asia Chengdu Regional Contest

 

Recommend

liuyiding   |   We have carefully selected several similar problems for you:  5508 5507 5506 5505 5504 

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI 3.1415926535897932384626#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   num<<1,le,mid#define rson    num<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;//const int maxn=    ;//const int maxm=    ;   ll n,m; bool use[12]; int num;  int a[12]; bool vis[10000+5]; struct Node {     ll yu;     int pre;     ll thi;     Node(){}     Node(ll yu,int pre,ll thi ):yu(yu),pre(pre),thi(thi){} } q[1000000];   void output(int x) {     if(x==-1) return;      output(q[x].pre);      printf("%lld",q[x].thi);  }bool  BFS() {     int r=0,f=0;     memset(vis,0,sizeof vis);     for(int i=1;i<=num;i++)     {         ll x=a[i];         if(x==0)  continue;         if(x%n==0)  {printf("%lld\n",x);return true;}         q[r++]= Node(x%n,-1,x );         vis[x%n]=1;     }     while(f<r)     {//         Node now=q.front();q.pop();       Node now=q[f++];         ll tmp=now.yu;         ll t;         for(int i=1;i<=num;i++)         {             ll x=a[i];             ll yu=(tmp*10+x)%n;              if(vis[yu])  continue;             q[r++]=Node(yu,f-1,x);             if(yu==0)  {output(r-1);putchar('\n');return true;}             vis[yu]=1;          }      }     return false;  }int main(){    int x,kase=0;    while(~scanf("%lld%d",&n,&m))    {        memset(use,0,sizeof use);        for(int i=1;i<=m;i++)        {            scanf("%d",&x);            use[x]=1;        }        num=0;        for(int i=0;i<=9;i++)        {            if(use[i])  continue;            num++;            a[num]=i;        }         printf("Case %d: ",++kase);        if(!BFS())  puts("-1");      }      return 0;}