DFS-POJ-1426-Find The Multiple

Find The Multiple
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

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

本来以为又是高精度又要搜索，结果只是一道范围在ULL以内的DFS水题。

////  main.cpp//  简单搜索-E-Find The Multiple////  Created by 袁子涵 on 15/8/1.//  Copyright (c) 2015年 袁子涵. All rights reserved.//#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int n;unsigned long long int m;int dfs(int step,unsigned long long int num){    if (num%n==0) {        m=num;        return 1;    }    if (step>=19) {        return 0;    }    if(dfs(step+1, num*10))        return 1;    if(dfs(step+1, num*10+1))        return 1;    return 0;}int main(int argc, const char * argv[]) {    while (scanf("%d",&n)!=EOF) {        if (n==0) {            break;        }        dfs(0,1);        cout << m << endl;    }    return 0;}