poj 1426 DFS

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12137		Accepted: 4993		Special Judge

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

---

给出一个数，求仅由0 和 1 组成的这个数的倍数；

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;

bool find;
int t,k;

void dfs(unsigned __int64 t,int n,int k)
{
    if(find)
        return;
    if(t%n==0)
    {
        find=true;
        printf("%I64u\n",t);
        return;
    }
    if(k==19)
        return ;
    dfs(t*10,n,k+1);
    dfs(t*10+1,n,k+1);
}


int main()
{
    int n;
    while(cin>>n,n)
    {
        find=false;
        dfs(1,n,1);
    }
    return 0;
}