POJ 1465 Multiple

Multiple

Time Limit: 1000MS Memory Limit: 32768KTotal Submissions: 6410 Accepted: 1386

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

223701211

Sample Output

1100

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <queue>#include <stack>#include <cmath>#include <algorithm>#define N 5020using namespace std;int n,m;int a[N];int f;struct node{    int digit;//数值    int yu;//余数    int pre;//用于递归输入结束判断    int id;//下标}que[N];bool b[N];void output(int id)//可能会出现大数所以递归输出{    if(que[id].pre==-1)    return;    output(que[id].pre);    printf("%d",que[id].digit);}void bfs(){    f=0;    memset(b,0,n);    int cnt=1,i,yu;    int id;    que[0].pre=-1;    que[0].yu=0;    que[0].id=0;    queue<int> Q;    Q.push(0);    while(!Q.empty())    {        id=Q.front();        Q.pop();        for(i=0;i<m;i++)        {            if(que[id].yu==0 &&a[i]==0)continue;            yu=(que[id].yu*10+a[i])%n;            if(!b[yu])            {                if(yu==0)                {                    output(id);                    printf("%d\n",a[i]);                    f=1;                    return;                }              b[yu]=1;              que[cnt].digit=a[i];              que[cnt].pre=id;              que[cnt].yu=yu;              que[cnt].id=cnt;              Q.push(cnt++);            }        }    }}int main(){    while(~scanf("%d",&n))    {        scanf("%d",&m);        for(int i=0;i<m;i++)        scanf("%d",&a[i]);        sort(a,a+m);        if(n==0)        {            printf("0\n");            continue;        }        bfs();        if(!f) printf("0\n");    }    return 0;}