POJ1426 Find The Multiple

题目链接：https://vjudge.net/problem/POJ-1426
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：对于一个数n，给出n的倍数m，要求m是由01组成的十进制数。
解题思路：运用DFS进行搜索，从1开始进行，随后在数的后面加1或者0，直到找到m为止。

Cpp Code

#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdio>
#define ll long long
using namespace std;
ll flag = 0,x,n;

void DFS(int step,ll x){
   if(step>19||flag==1){//long long二点最大值是19位，搜索19次找不到答案则退出
       return;
   }
   if(x%n==0){
       flag = 1;
       printf("%lld\n", x);
       return;
   }
   DFS(step + 1, x * 10);
   DFS(step + 1, x * 10+1);
}

int main(){
    while((scanf("%lld",&n)==1)&&n){
        flag = 0;
        DFS(1,1);
    }
    return 0;
}

Java Code

import java.util.Scanner;

public class Main {
	static long n;
	static boolean tag=false;
	public static void DFS(int step,long x) {
		if(step>19||tag==true) {//long的最大值是19位数，搜索19次找不到答案则退出
			return;
		}
		if(x%n==0) {
			tag=true;
			System.out.println(x);
			return;
		}
		DFS(step+1,x*10);
		DFS(step+1,x*10+1);
	}
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		while(true) {
			n=sc.nextLong();
			if(n==(long)0) {
				break;
			}
			tag=false;
			DFS(1,1);
		}
		sc.close();
	}
}