POJ - 1426 Find The Multiple（bfs）

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.

题意：求给出的这个数的只能由0,1组成的倍数。

思路：一开始是觉得直接bfs，m不超过100位数，但longlong也只能存19位，万一爆longlong怎么办 0.0 ，然后搜了一下题解，发现原来不会超，直接双入口的bfs就好了，其实我还是不知道为啥他们辣么肯定答案都在19位之内。。。 ，之后就简单了，一直往后搜*10，和*10+1就可以了，看到好多用dfs的，感觉还是广搜好些，毕竟短嘛，安全些2333，深搜还要规定在19位之内。

代码：

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <cctype>
#include <sstream>
#define INF 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef long long LL;
using namespace std;
const int maxn = 20 + 5;
const int mod = 1e8 + 7;

int n;
void bfs(int n){
queue<LL>q;
q.push(1);
while (!q.empty()){
LL 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() {
//freopen ("in.txt", "r", stdin);
while (~scanf ("%d",&n)&&n){
bfs(n);
}
return 0;
}