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

記錄餘數r，每一位的數dig，前驅pre，用陣列記錄一下餘數，如果餘數出現過，就不用擴充套件了，這裡要找最小的倍數，先把這些書排一下序，這題要輸出這個數，可以手寫佇列，便於儲存。另外，注意判斷一下n為0的情況,不然re,再就是注意輸出的格式

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 5050
int n,m;
int dig[15];
int flag[maxn];
struct node
{
    int dig;
    int r;
    int pre;
}q[maxn];
int bfs()
{
    memset(flag,0,sizeof(flag));
    int front=0,tail=1;
    q[front].pre=-1;
    q[front].dig=0;
    q[front].r=0;
    while(front<tail)
    {
        node node=q[front];
        int r=node.r;
        for(int i=0;i<m;i++)
        {
            int nr=(r*10+dig[i])%n;
            if(!flag[nr]&&(node.pre!=-1||dig[i]!=0))
            {
                flag[nr]=1;
                node.r=nr;
                node.dig=dig[i];
                node.pre=front;
                q[tail++]=node;
                if(nr==0)
                    return tail-1;
            }
        }
        front++;
    }
    return -1;
}
void print(int ans)
{
    if(ans>0)
    {
        print(q[ans].pre);
        printf("%d",q[ans].dig);
    }
}
int main()
{
    while(~scanf("%d",&n))
    {scanf("%d",&m);
        for(int i=0;i<m;i++)
         scanf("%d",&dig[i]);
         sort(dig,dig+m);
         if(n==0)
            {printf("0\n");
         continue;
            }
        int ans=bfs();
        if(ans==-1)
            printf("0\n");
            else
                {print(ans);
                printf("\n");
                }

    }
    return 0;
}