题目：
Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 35885 | Accepted: 14992 | 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
题意：找出一个只有0和1构成的，是n的倍数的数。
思路：由小到大依次bfs搜索，从1，10，11，100，101……一直搜下去。
没想明白的地方：搜索结果为啥不会爆long long呢。。不会数学证明啊！！！
还有为啥放到函数里输出就AC了，返回到main输出就WA掉了呢？
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<stack>typedef long long ll;using namespace std;void bfs(int n){queue<ll> q;q.push(1);while (!q.empty()){ll t = q.front();q.pop();if(t%n==0){cout<<t<<endl;return;}q.push(t*10);q.push(t*10 + 1);}
}int main(){ios::sync_with_stdio(false);int n;while(cin>>n && n){bfs(n);}return 0;
}