poj 1426 Find The Multiple

Find The Multiple
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 25716 Accepted: 10607 Special Judge
Description
Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.
Input
The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.
Output
For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.
Sample Input
2
6
19
0
Sample Output
10
100100100100100100
111111111111111111
Source
Dhaka 2002
题目大意
给出一个整数n，(1 <= n <= 200)。求出任意一个它的倍数m，要求m必须只由十进制的’0’或’1’组成。

/*
0和1 两种状态.
最高位必为1（除1外）. 
bfs余数最后状态余数为0 
每次扩展front()都会出现两种状态.
然后用BFS跑最优解. 
*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#define MAXN 1001
#define LL long long
using namespace std;
int  n;
bool b[MAXN];
LL bfs()
{
    memset(b,0,sizeof(b));
    queue<LL>qq;
    qq.push(1);
    b[1%n]=true;
    while(!qq.empty())
    {
        LL x=qq.front();
        if(x%n==0)
            return x;
        qq.pop() ;    
        int xx=x*10%n;
        if(!b[xx])
        {
            qq.push(x*10);
            b[xx]=true;

        }
        int yy=(x*10+1)%n;
        if(!b[yy])
        {
            qq.push(x*10+1);
            b[yy]=true;
        }
    }

}
int main()
{
    while(cin>>n)
    {
        if(!n) break;
         printf("%lld\n",bfs());
    }

    return 0;
}