TOJ 3031 Multiple

Description

a program that, given a natural number N between 0 and 4999 (inclusively), and M distinct decimal digits X1,X2..XM (at least one), finds the smallest strictly positive multiple of N that has no other digits besides X1,X2..XM (if such a multiple exists).

Input

The input has several data sets separated by an empty line, each data set having the following format:

On the first line - the number N
On the second line - the number M
On the following M lines - the digits X1,X2..XM.

Output

For
each data set, the program should write to standard output on a single
line the multiple, if such a multiple exists, and 0 otherwise.

An example of input and output:

Sample Input

22
3
7
0
1

2
1
1

Sample Output

110
0

Source

Southeastern Europe 2000

组队训练赛的一题，智商捉鸡。压根没头绪。看了别人的思路说是用广搜分别枚举每一位然后除以n是否模得0。

还有过程中可以对余数重复剪掉。

代码重新写了一遍。

 #include <stdio.h>
 #include <queue>
 #include <iostream>
 using namespace std;

 int n,m;
 int M[];
 bool visited[];

 struct Node{
     int digit;
     int pre;
     int mod;
     int cnt;
 }nod[];

 int bfs(){
     queue<int> Q;
     memset(visited,,sizeof(visited));
     int cur=;
     nod[cur].digit=;
     nod[cur].pre=-;
     nod[cur].mod=;
     nod[cur].cnt=cur;
     Q.push(cur++);
     while( !Q.empty() ){
         int now=Q.front();
         int now_mod;
         Q.pop();
         for(int i=; i<m; i++){
             if(nod[now].mod== && M[i]==)continue;
             now_mod=(nod[now].mod*+M[i])%n;
             if( !visited[now_mod] ){
                 visited[now_mod]=;
                 if(now_mod==){
                     int r[];
                     int index=;
                     r[index++]=M[i];
                     while( nod[now].pre!=- ){
                         r[index++]=nod[now].digit;
                         now=nod[now].pre;
                     }
                     for(int i=index-; i>=; i--){
                         printf("%d",r[i]);
                     }
                     printf("\n");
                     return ;
                 }else{
                     nod[cur].digit=M[i];
                     nod[cur].pre=nod[now].cnt;
                     nod[cur].mod=now_mod;
                     nod[cur].cnt=cur;
                     Q.push(cur++);
                 }
             }
         }
     }
     return ;
 }

 int main(int argc, char *argv[])
 {
     while( scanf("%d",&n)!=EOF ){
         scanf("%d",&m);
         for(int i=; i<m; i++){
             scanf("%d",&M[i]);
         }
         sort(M,M+m);
         if( n== || !bfs() ){
             puts("");
         }
     }
     return ;
 }