F - Find The Multiple POJ - 1426

2-Day2-F
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
题意：

给定一个正整数n，编写一个程序，找出n的非零倍数m，其十进制表示形式仅包含数字0和1。可以假设n不大于200，并且相应的m包含的十进制数字不超过100。

对于输入中的每个n值，打印一行，其中包含相应的m值。m的十进制表示形式不得包含超过100位的数字。如果给定的n值有多个解，则可以接受其中任何一个解。

思路：dfs即可

代码如下：

#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)
{
    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;
}