Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 51527 Accepted: 21543 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

题目大意：输出十进制数能够整除n的数（仅由1，0组成），开始以为是输出整除n任意一个二进制数。。。
比较简单的搜索，bfs，把求解的答案想象成一颗由0，1组成的完全二叉树，然后bfs一遍就行了。
有一个坑的地方，while(!st.empty())这个过不了，while(1)才能过。。。。也是看的评论才知道的。。。。不然一辈子过不了了
g++能过，c++过不了。。
代码：

#include<iostream>
#include<queue>
#include<cstdio> 
using namespace std;

typedef long long ll;
ll bfs(int n,ll s){
	queue<ll>st;
	st.push(s);
	while(1){
		ll now=st.front();st.pop();
		if(now%n==0){
			return now;
		}
		st.push((ll)now*10);
		st.push((ll)now*10+1);
	}
}
int main(){
	int n;
	while(scanf("%d",&n)==1&&n){
		cout<<bfs(n,1)<<endl;
	}
}