【题目描述】 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. 给定一个正整数n，请编写一个程序来寻找n的一个非零的倍数m，这个m用十进制表示时每一位上只包含0或者1。你可以假设n不大于200且m不多于100位。
【输入】 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. 输入包含多组数据，每组数据仅一行，只包含一个正整数n (1 <= n <= 200)。
【输出】 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. 对于输入的每组n，都输出任一符合条件的m。即使有多个符合条件的m，你也只需要输出一个即可。
【样例输入】 2 6 19 0
【样例输出】 10 100100100100100100 111111111111111111
题目链接：https://cn.vjudge.net/problem/POJ-1426
dfs，bfs均可，题干吓人，假装有100位，实际上long long就能过
代码如下：
#include <iostream> #include <queue> using namespace std; typedef unsigned long long ULL; int n; void bfs(ULL x) { queue<ULL> Q; Q.push(x); while(!Q.empty()) { ULL now=Q.front(); Q.pop(); if(now%n==0) { cout<<now<<endl; return; } Q.push(10ull*now); Q.push(10ull*now+1); } } int main() { std::ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0); while(cin>>n && n) bfs(1); return 0; }