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[101];
bool visited[6000];

struct Node{
    int digit;
    int pre;
    int mod;
    int cnt;    
}nod[6000];

int bfs(){
    queue<int> Q;
    memset(visited,0,sizeof(visited));
    int cur=0;
    nod[cur].digit=0;
    nod[cur].pre=-1;
    nod[cur].mod=0;
    nod[cur].cnt=cur;
    Q.push(cur++);
    while( !Q.empty() ){
        int now=Q.front();
        int now_mod;
        Q.pop();
        for(int i=0; i<m; i++){
            if(nod[now].mod==0 && M[i]==0)continue;
            now_mod=(nod[now].mod*10+M[i])%n;
            if( !visited[now_mod] ){
                visited[now_mod]=1;
                if(now_mod==0){
                    int r[6000];
                    int index=0;
                    r[index++]=M[i];
                    while( nod[now].pre!=-1 ){
                        r[index++]=nod[now].digit;
                        now=nod[now].pre;
                    }
                    for(int i=index-1; i>=0; i--){
                        printf("%d",r[i]);
                    }
                    printf("
");
                    return 1;
                }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 0;    
}

int main(int argc, char *argv[])
{
    while( scanf("%d",&n)!=EOF ){
        scanf("%d",&m);
        for(int i=0; i<m; i++){
            scanf("%d",&M[i]);
        }
        sort(M,M+m);
        if( n==0 || !bfs() ){
            puts("0");
        }
    }
    return 0;
}