E - Find The Multiple // dfs深度优先搜索（过程解释）总感觉存在问题，m不超过100位，用longlong就会存不开

本文介绍了一个编程问题：给定一个正整数n，如何找出一个只包含0和1的非零倍数m。通过深度优先搜索（DFS）和广度优先搜索（BFS）两种方法实现了这一目标，并提供了完整的C++代码示例。

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E - Find The Multiple

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

Sample Output

10
100100100100100100
111111111111111111

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
bool flag=false;
int n;
void dfs(int step,long long cst)
{
	// printf("cst=%lld\n",cst); 验证搜索过程 
	if(step==19 || flag)
	{
		return ;
	}
	if(cst%n==0)
	{
		flag=true;
		printf("%lld\n",cst);
		return ;
	}
	dfs(step+1,cst*10);
	dfs(step+1,cst*10+1);
}
int main()
{
	while(scanf("%d",&n),n)
	{
		flag=false;
		dfs(0,1);
	}
	return 0;
}

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
int n;
void bfs(long long int cst)
{
	queue<long long>Q;
	Q.push(cst);
	while(Q.size())
	{
		long long y=Q.front();
		Q.pop();
		if(y%n==0)
		{
			printf("%lld\n",y);
			return ;
		}
		Q.push(y*10);
		Q.push(y*10+1);
	}
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
		  break;
		bfs(1);
	}
	return 0;
}