POJ

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|1typedef 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;}