POJ 1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21832		Accepted: 8975		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

题目传送门：POJ1426 Find The Multiple

题目挺简单的，就是输入一个数n，让你求一个数能整除n。但是这个数只能由0和1组成。

思路就是BFS来破咯，设需要求的这个数为x。x从1开始，如果能整除n就输出，不能的话。分别求出x=x*10，x=x*10+1，也就是在x后填了1或0，再去看能不能整除。还不能整除就再重复上述操作。挺简单的吧~  题目上面写着 Special Judge，说明答案不止一个，输出一个符合题目要求的就ok啦~ 下面是AC代码【本题交G++，交C++超时】

#include <cstdio>
#include <iostream>
#include <queue>
using namespace std;
void bfs(int n)
{
    queue<long long>Q;
    Q.push(1);
    while(!Q.empty())
    {
        long long x;
        x=Q.front();
        Q.pop();
        if(x%n==0)
        {
            printf("%lld\n",x);
            return ;
        }
        Q.push(x*10);
        Q.push(x*10+1);
    }
}

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