ural 1495 bfs

定义状态visit[i][j]表示最后一位是i且该数字mod n的余数是j的情况，然后按位来bfs即可，转移时先考虑1再考虑2，这样求出来的答案就是长度和数值都最小的了。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 1000007;
int n, head, tail;
bool visit[3][N];

struct Node
{
    int d, r, len, pre;
    Node(){}
    Node( int _d, int _r, int _len, int _pre )
    {
        d = _d, r = _r, len = _len, pre = _pre;
    }
} q[N << 1];

void dfs( int x )
{
    if ( x == -1 ) return ;
    dfs( q[x].pre );
    printf("%d", q[x].d);
}

void bfs()
{
    memset( visit, 0, sizeof(visit) );
    head = tail = 0;
    q[tail++] = Node( 1, 1 % n, 1, -1 );
    q[tail++] = Node( 2, 2 % n, 1, -1 );
    visit[1][1 % n] = 1;
    visit[2][2 % n] = 1;
    while ( head < tail )
    {
        if ( q[head].len > 30 )
        {
            break;
        }
        if ( q[head].r == 0 )
        {
            dfs(head);
            putchar('
');
            return ;
        }
        for ( int k = 1; k <= 2; k++ )
        {
            int rr = ( q[head].r * 10 + k ) % n;
            if ( !visit[k][rr] )
            {
                q[tail++] = Node( k, rr, q[head].len + 1, head );
                visit[k][rr] = 1;
            }
        }
        head++;
    }
    printf("Impossible
");
}

int main ()
{
    while ( scanf("%d", &n) != EOF )
    {
        bfs();
    }
    return 0;
}