Zju1136 Multiple N的倍数

1289: Zju1136 Multiple N的倍数

Time Limit: 1 Sec  Memory Limit: 256 MB
Submit: 62  Solved: 46
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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

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.

Sample Input

22
3
7
0
1

Sample Output

110

HINT

 

Source

#include <iostream>
using namespace std;
int num[10000],y[10000],pp[10000];
bool o[5000];
int f[10];
int m,n,i,j,k,temp,head,tail;

void output(int head)
{
    if (pp[head]!=0) 
    output (pp[head]);
    cout<<num[head];
}
void bfs()
{
    head=0; 
    tail=0; 
    k=0;
    do
    {
        for (i=1;i<=m;i++)
        {
            if (head==0 && f[i]==0) continue;
            if (o[(y[head]*10+f[i])%n]) continue;
            tail++; 
            y[tail]=(y[head]*10+f[i])%n;
            pp[tail]=head; 
            num[tail]=f[i]; 
            o[y[tail]]=true;
            if (y[tail]==0)
            {
                output(tail); 
                k=1; 
                break;
            }
        }
        if (k==1) break; 
        head++;
    } while(head<=tail);
}
int main()
{
        cin>>n>>m;
        for (i=1;i<=m;i++)
        cin>>f[i];
        if (n==0)
        {
            cout<<0<<endl;
            return 0;
        }
        for (i=1;i<=m-1;i++)
        for (j=i+1;j<=m;j++)
        if (f[j]<f[i])
        {
            temp=f[j]; 
            f[j]=f[i]; 
            f[i]=temp;
        }
        bfs();
        if (k==0)
        cout<<0<<endl;
        cout<<endl;
    return 0;
}