poj2352，poj 1426 BFS

一个数n，求一个数m，m由0和1构成，且是n的倍数，若存在多个可任意输出一个即可

数字不会超过19位，用long long int保存。

#pragma warning(disable:4996)
#include<iostream>
#include<string>
#include<cmath>
#include<ctype.h>
#include<memory.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<iomanip>
#include<set>
#include<list>
#include<vector>
#include<stack>
#include<queue>
#define ll long long int
using namespace std;
queue<ll> q;
ll bfs(int n)
{while (!q.empty())q.pop();q.push(1);while (!q.empty()){ll t = q.front();q.pop();if (t % n == 0)//是n的倍数return t;q.push(t * 10);// 10q.push(t * 10 + 1);// 11}return 0;
}
int n;
int main()
{while (1){cin >> n;if (n == 0)break;cout << bfs(n) << endl;}return 0;
}